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ge of a subspace of the Cantor Space (G, τ ). Further, if (X, τ 1) is compact, then the subspace can be chosen to be closed in (G, τ ). Let φ be the continuous mapping of (G, τ ) onto I∞ shown to exist in the Proof. proof of Theorem 9.3.8. By Urysohn’s Theorem 9.4.19, (X, τ 1) is homeomorphic to a subspace (Y, τ 2) of I∞. Let the homeomorphism of (Y, τ 2) onto (X, τ 1) be Θ. Let Z = ψ−1(Y ) and τ 3 be the subspace topology on Z. Then Θ ◦ ψ is a continuous mapping of (Z, τ 3) onto (X, τ 1). So (X, τ 1) is a continuous image of the subspace (Z, τ 3) of (G, τ ). Further if (X, τ 1) is compact, then (Y, τ 2) is compact and hence closed in I∞. So Z = ψ−1(Y ) is a closed subset of (G, τ ), as required. Let (Y, τ 1) be a (non-empty) closed subspace of the 9.5.3 Proposition. Cantor Space (G, τ ). Then there exists a continuous mapping of (G, τ ) onto (Y, τ 1). Let (G, τ ) be the set of all real numbers which can be written in the Proof. ai form ∞ 6i , where each ai = 0 or 5, with the subspace topology induced from R. i=1 The space (G, τ ) is called the middle two-thirds Cantor Space. Clearly (G, τ ) is homeomorphic to the Cantor Space (G, τ ). We can regard (Y, τ 1) as a closed subspace of (G, τ ) and seek a continuous mapping of (G, τ ) onto (Y, τ 1). Before proceeding, observe from the construction of the middle two thirds Cantor space that if g1 ∈ G and g2 ∈ G, then /∈ G. The map ψ : (G, τ ) −→ (Y, τ 1) which we seek is deﬁned as follows: for g ∈ G, ψ(g) is the unique element of Y which is closest to g in the euclidean metric on R. However we have to prove that such a unique closest element exists. g1+g2 2 Fix g ∈ G. Then the map dg : (Y, τ 1) −→ R given by dg(y) = \|g − y\| is continuous. As (Y, τ 1) is compact, Proposition 7.2.15 implies that dg(Y ) has a least element. So there exists an element of (Y, τ 1) which is closest to g. Suppose there are two such elements y1 and y2 in Y which are equally close to g. Then g = y1+y2 /∈ G, . But y1 ∈ G and y2 ∈ G and so, as observed above, g = y1+y2 2 2 9.5. PEANO’S THEOREM 249 which is a contradiction. So there exists a unique element of Y which is closest to g. Call this element ψ(g). It is clear that the map ψ : (G, τ ) −→ (Y, τ 1) is surjective, since for each y ∈ Y , ψ(y) = y. To prove continuity of ψ, let g ∈ G. Let ε be any given positive real number. Then it suﬃces, by Corollary 6.2.4, to ﬁnd a δ > 0, such that if x ∈ G and \|g − x\| < δ then \|ψ(g) − ψ(x)\| < ε. Consider ﬁrstly the case when g ∈ Y , so ψ(g) = g. Put δ = ε 2. Then for x ∈ G with \|g − x\| < δ we have \|ψ(g) − ψ(x)\| = \|g − ψ(x)\| \|x − ψ(x)\| + \|g − x\| \|x − g\| + \|g − x\|, by deﬁnition of ψ since g ∈ Y = 2\|x − g\| < 2δ = ε, as required. Now consider the case when g /∈ Y , so g = ψ(g). Without loss of generality, assume ψ(g) < g and put a = g − ψ(g). 2, g + a If the set Y ∩ [g, 1] = Ø, then ψ(x) = ψ(g) for all x ∈ (g − a 2). Thus for δ < a 2, we have \|ψ(x) − ψ(g)\| = 0 < ε, as required. If Y ∩ [g, 1] = Ø, then as Y ∩ [g, 1] is compact it has a least element y > g. Indeed by the deﬁnition of ψ, if b = y − g, then b > a. Now put δ = b−a 2 . So if x ∈ G with \|g − x\| < δ, then either ψ(x) = ψ(g) or ψ(x) = y. Observe that \|x − ψ(g)\| \|x − g\| + \|g − ψ(g)\| < while So ψ(x) = ψ(g). \|x − y\| \|g − y\| − \|g − x . Thus \|ψ(x) − ψ(g)\| = 0 < ε, as required. Hence ψ is continuous. Thus we obtain from Propositions 9.5.2 and 9.5.3 the following theorem of Alexandroﬀ and Urysohn: 250 CHAPTER 9. COUNTABLE PRODUCTS 9.5.4 Theorem. Every compact metrizable space is a continuous image of the Cantor Space. 9.5.5 Remark. The converse of Theorem 9.5.4 is false. It is not true that every continuous image of a Cantor Space is a compact metrizable space. (Find an example.) However, an analogous statement is true if we look only at Hausdorﬀ spaces. Indeed we have the following proposition. Let f be a continuous mapping of a compact metric 9.5.6 Proposition. space (X, d) onto a Hausdorﬀ space (Y, τ 1). Then (Y, τ 1) is compact and metrizable. Proof. Since every continuous image of a compact space is compact, the space (Y, τ 1) is certainly compact. As the map f is surjective, we can deﬁne a metric d1 on Y as follows: for each y1, y2 ∈ Y , d1(y1, y2) equals inf n∈N {d(a1, b1) + · · · + d(an, bn) : f (a1) = y1, f (bn) = y2, bi = ai+1, i = 1, . . . , n − 1}. We need to show that d1 is indeed a metric; this is left as an exercise. (See Exercises 9.5 #3.) Let τ 2 be the topology induced on Y by d1. We have to show that τ 1 = τ 2. Firstly, by the deﬁnition of d1, f : (X, τ ) −→ (Y, τ 2) is certainly continuous and so (X, τ 2) is compact. Observe that for a subset C of Y , C is a closed subset of (Y, τ 1) ⇒ f −1(C) is a closed subset of (X, τ ) ⇒ f −1(C) is a compact subset of (X, τ ) ⇒ f (f −1(C)) is a compact subset of (Y, τ 2) ⇒ C is a compact subset of (Y, τ 2) ⇒ C is closed in (Y, τ 2). So τ 1 ⊆ τ 2. Similarly we can prove τ 2 ⊆ τ 1, and thus τ 1 = τ 2. 9.5. PEANO’S THEOREM 251 9.5.7 Corollary. Let (X, τ ) be a Hausdorﬀ space. Then it is a continuous image of the Cantor Space if and only if it is compact and metrizable. Finally in this chapter we turn to space-ﬁlling curves. 9.5.8 Remark. Everyone thinks he or she knows what a “curve” is. Formally we can deﬁne a curve in R2 to be the set f [0, 1], where f is a continuous map f : [0, 1] −→ R2. It seems intuitively clear that a curve has no breadth and hence zero area. This is false! In fact there exist space-ﬁlling curves; that is, f (I) has nonIndeed the next theorem shows that there exists a continuous mapping zero area. of [0, 1] onto the product space [0, 1] × [0, 1]. 9.5.9 Theorem. (Peano) continuous mapping ψn of [0, 1] onto the n-cube In. For each positive integer n, there exists a By Theorem 9.5.4, there exists a continuous mapping φn of the Cantor Proof. Space (G, τ ) onto the n-cube In. As (G, τ ) is obtained from [0, 1] by successively dropping out middle thirds, we extend φn to a continuous mapping ψn : [0, 1] −→ In by deﬁning ψn to be linear on each omitted interval; that is, if (a, b) is one of the open intervals comprising [0, 1] \ G, then ψn is deﬁned on (a, b) by ψn (αa + (1 − α) b) = α φn(a) + (1 − α) φn(b), 0 α 1. It is easily veriﬁed that ψn is continuous. We conclude this chapter by stating (but not proving) the Hahn-Mazurkiewicz Theorem which characterizes those Hausdorﬀ spaces which are continuous images of [0,1]. [For a proof of the theorem see Wilder [424] and p. 221 of Kuratowski [249].] But ﬁrst we need a deﬁnition. 252 CHAPTER 9. COUNTABLE PRODUCTS 9.5.10 Deﬁnition. A topological space (X, τ ) is said to be locally connected if it has a basis of connected (open) sets. Every discrete space is locally connected as are Rn and Sn, for 9.5.11 Remark. all n 1. However, not every connected space is locally connected. (See Exercises 8.4 #6.) Our ﬁnal theorem in this section is beautiful. It was proved by Hans Hahn (1879– 1934) and Stefan Mazurkiewicz (1888–1945). Its proof can be found in Hocking and Young [189]. (Hahn-Mazurkiewicz Theorem) Let (X, τ ) be a 9.5.12 Theorem. Hausdorﬀ space. Then (X, τ ) is a continuous image of [0, 1] if and only if it is compact, connected, metrizable and locally connected. Hahn Mazurkiewicz 9.5. PEANO’S THEOREM 253 Exercises 9.5 1. Let S ⊂ R2 be the set of points inside and on the triangle ABC, which has a right angle at A and satisﬁes AC > AB. This exercise outlines the construction of a continuous surjection f : [0, 1] → S. (This provides an easy example of a space-ﬁlling curve; the curve f ([0, 1]) ﬁlls S.) A ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ...................................................................................................................................................................................................................................................................................................................................................... E F .................................................................................................................................................................... G H ............................................................................... I D C B Let D on BC be such that AD is perpendicular to BC. Let a = ·a1a2a3 . . . be a binary decimal, so that each an is 0 or 1. Then we construct a sequence of points of S as follows : E is the foot of the perpendicular from D onto the hypotenuse of the larger or smaller of the triangles ADB, ADC according as a1 = 1 or 0, respectively. This construction is now repeated using E instead of D and the appropriate triangle of ADB, ADC instead of ABC. For example, the ﬁgure above illustrates the points E to I for the binary decimal .11001 . . . . Give a rigorous in
ductive deﬁnition of the sequence of points and prove (i) the sequence of points tends to a limit L(a) in S; (ii) if λ ∈ [0, 1] is represented by distinct binary decimals a, a then L(a) = L(a); hence, the point L(λ) in S is uniquely deﬁned; (iii) if f : [0, 1] → S is given by f (λ) = L(λ), then f is surjective; (iv) f is continuous. 254 CHAPTER 9. COUNTABLE PRODUCTS 2. Let (G, τ ) be the Cantor Space and consider the mappings where and φi : (G, τ ) → [0, 1], i = 1, 2, φ1 φ2 ∞ i=1 ∞ i=1 ai 3i ai 3i = a1 22 + a3 23 + · · · + a2n−1 2n+1 + . . . = a2 22 + a4 23 + · · · + a2n 2n+1 + . . . . (i) Prove that φ1 and φ2 are continuous. (ii) Prove that the map a → φ1(a), φ2(a) is a continuous map of (G, τ ) onto [0, 1] × [0, 1]. (iii) If a and b ∈ (G, τ ) and (a, b) ∩ G = Ø, deﬁne, for j = 1, 2 , φj(x) = b − x b − a φj(a) + x − ab − a φj(b), a x b. Show that x → φ1(x), φ2(x) is a continuous mapping of [0, 1] onto [0, 1] × [0, 1] and that each point of [0, 1] × [0, 1] is the image of at most three points of [0, 1]. 3. Prove that d1 in the proof of Proposition 9.5.6 is indeed a metric. 9.6 Postscript In this chapter we have extended the notion of a product of a ﬁnite number of topological spaces to that of the product of a countable number of topological spaces. While this step is a natural one, it has led us to a rich collection of results, some of which are very surprising. We proved that a countable product of topological spaces with property P has (iii) property P, where P is any of the following: Hausdorﬀ (iv) metrizable (v) connected (vi) totally disconnected (vii) second It is also true when P is compact, this result being the Tychonoﬀ (ii) T1-space (i) T0-space countable. 9.6. POSTSCRIPT 255 Theorem for countable products. The proof of the countable Tychonoﬀ Theorem for metrizable spaces presented here is quite diﬀerent from the standard one which appears in the next section. Our proof relies on the Cantor Space. The Cantor Space was deﬁned to be a certain subspace of [0, 1]. Later it was shown that it is homeomorphic to a countably inﬁnite product of 2-point discrete spaces. The Cantor Space appears to be the kind of pathological example pure mathematicians are fond of producing in order to show that some general statement is false. But it turns out to be much more than this. The Alexandroﬀ-Urysohn Theorem 9.5.4 says that every compact metrizable space is an image of the Cantor Space. In particular [0, 1] and the Hilbert cube (a countable inﬁnite product of copies of [0, 1]) is a continuous image of the Cantor Space. This leads us to the existence of space-ﬁlling curves – in particular, we show that there exists a continuous map of [0, 1] onto the cube [0, 1]n, for each positive integer n. We stated, but did not prove, the Hahn-Mazurkiewicz Theorem 9.5.12: The Hausdorﬀ space (X, τ ) is an image of [0, 1] if and only if it is compact connected locally connected and metrizable. Next we mention Urysohn’s Theorem 9.4.19, which says that a space is separable and metrizable if and only if it is homeomorphic to a subspace of the Hilbert cube. This shows that [0, 1] is not just a “nice” topological space, but a “generator” of the important class of separable metrizable spaces via the formation of subspaces and countable products. Finally we mention a beautiful and deep theorem related to countably inﬁnite products which we have not proved in this chapter. For proofs of this theorem and discussion, see Anderson and Bing [12],Toru`nczyk [396], and Bessaga and Pelczynski [41]. We include a few related and surprising results from Bessaga and Pelczynski [41]; for example, the open unit ball, the closed unit ball, the unit sphere, any non-empty open convex subset, and any closed convex set with non-empty interior in a separable inﬁnite-dimensional Banach space B are homeomorphic to each other and to the whole space B. Note that, within the context of topological vector spaces a Frechet space is a complete metrizable locally convex topological vector space. 256 CHAPTER 9. COUNTABLE PRODUCTS 9.6.1 Theorem. (Anderson-Bessaga-Kadec-Pelczyński-Toru`nczyk) (i) Every separable inﬁnite-dimensional Fréchet space is homeomorphic to the countably inﬁnite product Rℵ0; (2) Every inﬁnite-dimensional Fréchet space is homeomorphic to a Hilbert space. 9.6.2 Corollary. homeomorphic to the countably inﬁnite product Rℵ0. Every separable inﬁnite-dimensional Banach space is 9.6.3 Corollary. homeomorphic to the separable Hilbert space 2. Every separable inﬁnite-dimensional Banach space is 9.6.4 Corollary. spaces, then B1 is homeomorphic to B2. If B1 and B2 are any separable inﬁnite-dimensional Banach 9.6.5 Theorem. (Bessaga and Pelczynski [41, Chapter VI, Theorem 6.2]) Let B be an inﬁnite-dimensional Banach space, S = {x : x ∈ B, \|\|x\|\| = r} a sphere in B of radius r > 0, and C = {x : x ∈ B, \|\|X\|\| r\|\|} a closed ball in B. Then B, S and C are homeomorphic. Let N be an inﬁnite-dimensional normed vector space and is 9.6.6 Remark. O = {x : \|\|x\|\| < 1} an open ball in N of radius 1. Observe that x → x a continuous map of N onto O with continuous inverse x → x . So the open unit ball O is homeomorphic to N . Indeed every open ball in N is homeomorphic to N . 1−\|\|x\|\| 1+\|\|x\|\| 9.6. POSTSCRIPT 257 9.6.7 Corollary. (Bessaga and Pelczynski [41, Chapter VI, Theorem 6.2]) If S = {x : x ∈ B, \|\|x\|\| = r} Let B be an inﬁnite-dimensional Banach space. a sphere in B of radius r > 0, C = {x : x ∈ B, \|\|X\|\| r\|\|} a closed ball in B, and O = {x : x ∈ B, \|\|x\|\| < r} an open ball in B, then B, S, C and O are homeomorphic. Indeed, if E is a closed convex subspace of an inﬁnite-dimensional Fréchet space F such that E has non-empty interior, then E is homeomorphic to F . If F is a separable inﬁnite-dimensional Fréchet space, B is 9.6.8 Corollary. a separable inﬁnite-dimensional Banach space, S is a sphere in B, C is a closed ball in B, O is an open ball in B, W is an open convex subspace of F , E is a Gδ subspace of F with non-empty interior, then the following topological spaces are homeomorphic. (a) Rℵ0; (b) (Rℵ0)m, where m is any positive integer or ℵ0; (c) 2; (d) (2)m, where m is any positive integer or ℵ0; (e) F ; (f ) F m, where m is any positive integer or ℵ0; (g) B; (h) Bm, where m is any positive integer or ℵ0; (i) Gi, where each Gi is a separable inﬁnite-dimensional Fréchet or Banach ∞ i=1 or Hilbert space; (j) Sm, where m is any positive integer or ℵ0; (k) Cm, where m is any positive integer or ℵ0; (l) Om, where m is any positive integer or ℵ0; (m) W m, where m is any positive integer or ℵ0; (n) Em, where m is any positive integer or ℵ0. 258 CHAPTER 9. COUNTABLE PRODUCTS 9.6.9 Remark. Teachers of topology and authors of books on topology should give some thought to Corollary 9.6.8. Often when teaching topology some teachers give many examples of topological spaces drawn from the set of inﬁnite-dimensional separable Banach spaces. But we now see that these spaces are all homeomorphic, so as far as topology is concerned they represent the same example over and over again. 9.7 Credits for Images 1. Hans Hahn. Copyright information: http://www-history.mcs.st-andrews.ac.uk/Miscellaneous/Copyright.html. 2. Stefan Mazurkiewicz. This photograph is in the public domain because according to the Art. 3 of copyright law of March 29, 1926 of the Republic of Poland and Art. 2 of copyright law of July 10, 1952 of the People’s Republic of Poland, all photographs by Polish photographers (or published for the ﬁrst time in Poland or simultaneously in Poland and abroad) published without a clear copyright notice before the law was changed on May 23, 1994 are assumed public domain in Poland. Chapter 10 Tychonoﬀ ’s Theorem Introduction In Chapter 9 we deﬁned the product of a countably inﬁnite family of topological spaces. We now proceed to deﬁne the product of any family of topological spaces by replacing the set {1, 2, . . . , n, . . . } by an arbitrary index set I. The central result will be the general Tychonoﬀ Theorem. The reader should be aware that this chapter is more sophisticated and challenging than previous chapters. However, the reward is that you will experience, and hopefully enjoy, some beautiful mathematics. Andrey Nikolayevich Tychonoﬀ (or Tikhonov) Russian mathematician (1906–1993) 259 260 CHAPTER 10. TYCHONOFF’S THEOREM 10.1 The Product Topology For All Products 10.1.1 Deﬁnitions. Let I be a set, and for each i ∈ I, let (Xi, τ i) be a topological space. We write the indexed family of topological spaces as {(Xi, τ i) : i ∈ I}. Then the product (or cartesian product) of the family of sets {Xi : i ∈ I} is denoted by i∈I Xi, and consists of the set of all functions f : I −→ i∈I Xi such that fi = xi ∈ Xi. We denote the element f of the product by i∈I xi, and refer to f (i) = xi as the ith coordinate. If I = {1, 2} then i∈{1,2} Xi is just the set of all functions f : {1, 2} → X1 ∪ X2 such that f (1) ∈ X1 and f (2) ∈ X2. A moment’s thought shows that i∈{1,2} Xi is a set “isomorphic to” X1 × X2. Similarly if I = {1, 2, . . . , n, . . . }, then i∈I Xi is “isomorphic to” our previously deﬁned ∞ i=1 Xi. The product space, denoted by i∈I Xi with the topology τ having as its basis the family i∈I(Xi, τ i), consists of the product set B = i∈I Oi : Oi ∈ τ i and Oi = Xi, for all but a ﬁnite number of i . The topology τ is called the product topology (or the Tychonoﬀ topology). Although we have deﬁned i∈I (Xi, τ i) rather diﬀerently to 10.1.2 Remark. the way we did when I was countably inﬁnite or ﬁnite you should be able to convince yourself that when I is countably inﬁnite or ﬁnite the new deﬁnition is equivalent to our previous ones. Once this is realized many results on countable products can be proved for uncountable products in an analogous fashion. We state them below. It is left as an exercise for the reader to prove these results for uncountable products. 10.1. THE PRODUCT TOPOLOGY FOR ALL PRODUCTS 261 10.1.3 Proposition. subset of a topological space (X, τ i). Then Let I be a set an
d for i ∈ I, let Ci be a closed i∈I Ci is a closed subset of i∈I (Xi, τ i). 10.1.4 Proposition. a family of topological spaces having product space ( i ∈ I, Bi is a basis for τ i, then Let I be a set and and let {(Xi, τ i) : i ∈ I} be If for each i∈I Xi, τ ). Oi : Oi ∈ Bi and Oi = Xi for all but a ﬁnite number of i B = i∈I is a basis for τ . 10.1.5 Proposition. of topological spaces having product space ( pj : each i∈I Xi −→ Xj be the projection mapping; that is, pj( i∈I xi ∈ Let I be a set and let {(Xi, τ i) : i ∈ I} be a family i∈I Xi, τ ). For each j ∈ I, let i∈I xi) = xj, for i∈I Xi. Then (i) each pj is a continuous surjective open mapping, and (ii) τ is the coarsest topology on the set i∈I Xi such that each pj continuous. is 10.1.6 Proposition. of topological spaces with product space homeomorphic to a subspace of Let I be a set and let {(Xi, τ i) : i ∈ I} be a family i∈I (Xi, τ i). Then each (Xi, τ i) is i∈I (Xi, τ i). 262 CHAPTER 10. TYCHONOFF’S THEOREM i) : i ∈ I} be families of topological spaces. If hi : (Xi, τ i) −→ (Yi, τ 10.1.7 Proposition. {(Yi, τ a continuous mapping, for each i ∈ I, then h : is continuous, where h( Let I be a set and let {(Xi, τ i) : i ∈ I} and i) is i∈I (Yi, τ i) i∈I (Xi, τ i) −→ i∈I xi) = i∈I hi(xi). Let I be a set and let {(Xi, τ i) : i ∈ I} be a 10.1.8 Proposition. family of topological spaces and f a mapping of a topological space (Y, τ ) i∈I (Xi, τ i). Then f is continuous if and only if each mapping pi ◦ f : into (Y, τ ) −→ (Xi, τ i) is continuous, where pi, i ∈ I, denotes the projection mapping of i∈I (Xi, τ i) onto (Xi, τ i). Let I be an index set 10.1.9 Lemma. (The Embedding Lemma) and {(Yi, τ i) : i ∈ I} a family of topological spaces and for each i ∈ I, let fi be a mapping of a topological space (X, τ ) into (Yi, τ i). Further let e : (X, τ ) −→ i∈I fi(x), for all x ∈ X. Then e is a homeomorphism of (X, τ ) onto the space (e(X), τ ), where τ is the subspace topology on e(X) if i∈I (Yi, τ i) be the evaluation map; that is, e(x) = (i) each fi is continuous. (ii) the family {fi : i ∈ I} separates points of X; that is, if x1 and x2 are in X with x1 = x2, then for some i ∈ I, fi(x1) = f1(x2), and (iii) the family {fi : i ∈ I} separates points and closed sets; that is, for x ∈ X and A any closed subset of (X, τ ) not containing x, fi(x) /∈ fi(A), for some i ∈ I. 10.1.10 Corollary. (ii) is superﬂuous. If (X, τ ) in Lemma 10.1.9 is a T1-space, then condition 10.1. THE PRODUCT TOPOLOGY FOR ALL PRODUCTS 263 Let (X, τ ) and (Y, τ ) be topological spaces. Then 10.1.11 Deﬁnitions. we say that (X, τ ) can be embedded in (Y, τ ) if there exists a continuous mapping f : (X, τ ) −→ (Y, τ ), such that f : (X, τ ) −→ (f (X), τ ) is a homeomorphism, where τ is the subspace topology on f (X) from (Y, τ ). The mapping f : (X, τ ) −→ (Y, τ ) is said to be an embedding. Exercises 10.1 1. For each i ∈ I, some index set, let (Ai, τ i) is a subspace of i) be a subspace of (Xi, τ i). i∈I (Xi, τ i). (i) Prove that (ii) Prove that (iii) Prove that Int( i∈I (Ai, τ i∈I Ai = i∈I Ai . i∈I Ai) ⊆ (iv) Give an example where equality does not hold in (iii). i∈I (Int(Ai)). 2. Let J be any index set, and for each j ∈ J, (Gj, τ j) a topological space homeomorphic to the Cantor Space, and Ij a topological space homeomorphic j∈J Ij is a continuous image of to [0, 1]. Prove that 3. Let {(Xj, τ j) : j ∈ J} be any inﬁnite family of separable metrizable spaces. j∈J (Xj, τ j) is homeomorphic to a subspace of j , where j∈J (Gj, Tj). j∈J I∞ is homeomorphic to the Hilbert cube. Prove that each I∞ j 264 CHAPTER 10. TYCHONOFF’S THEOREM 4. (i) Let J be any inﬁnite index set and {(Xi,j, τ i,j) : i ∈ N and j ∈ J} a family of homeomorphic topological spaces. Prove that (Xi,j, τ i,j) ∼= (X1,j, τ 1,j). j∈J i∈N (ii) For each j ∈ J , any inﬁnite index set, let (Aj, τ j) be homeomorphic to the discrete space {0, 2} and (Gj, Tj) homeomorphic to the Cantor Space. Deduce from (i) that j∈J (Aj, τ j) ∼= j∈J j∈J (Gj, Tj). (iii) For each j ∈ J , any inﬁnite index set, let Ij be homeomorphic to [0, 1], and j homeomorphic to the Hilbert cube I∞. Deduce from (i) that I∞ ∼= Ij j∈J I∞ j . j∈J (iv) Let J, Ij, I∞ j∈J I∞ j , and (Aj, τ j) be as in (ii) and (iii). Prove that j∈J Ij and j are continuous images of j∈J (Aj, τ j). (v) Let J and Ij be as in (iii). metrizable space, deduce from #3 above and (iii) above that is homeomorphic to a subspace of If, for each j ∈ J , (Xj, τ j) is a separable j∈J (Xj, τ j) j∈J Ij. 5. If I is an index set and each (Vi, τ i), i ∈ I, is a topological vector space, then i∈I Vi, τ ), with the obvious vector space structure and the product topology ( τ , is a topological vector space. Deduce that if each (Vi, τ i) is a locally convex space (or more particularly a normed vector space or a Banach space) then ( i∈I Vi, τ ) is a locally convex space. 10.2. ZORN’S LEMMA 265 6.** Let (E, τ ) be a Hausdorﬀ locally convex space. Then there exists an index set I, a set {Bi : i ∈ I} of Banach spaces, and a linear map Φ : E → i∈I Bi (with the product topology) which is an embedding. (In other words, every Hausdorﬀ locally convex space can be embedded as a topological vector subspace of a product of Banach spaces.) [Hint. Let {pi : i ∈ I} be the set of all continuous seminorms on (E, τ ). Verify that for each i ∈ I, Ai = {x : x ∈ E, pi(x) = 0} is a vector subspace of E. Let Ni be the quotient vector space E/Ai. Deﬁne a seminorm qi on Ni = E/Ai in a natural way and verify that it is in fact a norm. Thus Ni is a normed vector space. Using Exercises 6.3 #12, let Bi be the Banach space which is the completion of the normed vector space Ni. Verify that the natural linear maps φ : E → Bi are continuous. Prove that these linear maps φi, i ∈ I, deﬁne an embedding linear map Φ : E → i∈I Bi.] 10.2 Zorn’s Lemma Our next task is to prove the general Tychonoﬀ Theorem which says that any product of compact spaces is compact. However, to do this we need to use Zorn’s Lemma which requires a little preparation. (Davey and Priestley [97]) A partial order on a set 10.2.1 Deﬁnitions. X is a binary relation, denoted by , which has the properties: (i) x x, for all x ∈ X (reﬂexive) (ii) if x y and y x, then x = y, for x, y ∈ X (antisymmetric), and (iii) if x y and y z, then x z, for x, y, z ∈ X (transitive) The set X equipped with the partial order is called a partially ordered set or a poset and is denoted by (X, ). If x y and x = y, then we write x < y. 10.2.2 Examples. The prototype of a partially ordered set is the set N of all natural numbers equipped with the usual ordering of natural numbers. Similarly the sets Z, Q, and R with their usual orderings form partially ordered sets. 266 CHAPTER 10. TYCHONOFF’S THEOREM 10.2.3 Example. Let N be the set of natural numbers and let be deﬁned as follows: n m if n divides m So 3 6 but 3 5. (It is left as an exercise to verify that with this ordering N is a partially ordered set.) 10.2.4 Example. a partial ordering on X by putting Let X be the class of all subsets of a set U . We can deﬁne A B if A is a subset of B where A and B are in X. It is easily veriﬁed that this is a partial order. 10.2.5 Example. partial order ∗ on X by deﬁning Let (X, ) be a partially ordered set. We can deﬁne a new x ∗ y if y x. 10.2. ZORN’S LEMMA 267 10.2.6 Example. There is a convenient way of picturing partially ordered sets; this is by an order diagram. i • f g h •...........................................................................................................................................................................................• ..............................................................................................................................................................................................................................................................• ...................................................................................................................................................................................•e ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................... • b • c j • ........................................................................................................................................................................................... ................................................................................................................................................................... • a • d An element x is less than an element y if and only if one can go from x to y by moving upwards on line segments. So in our order diagram a < b, a < g, a < h, a < i, a < j, a < f, b < g, b < h, b < i, b < f, c < b, c < f, c < g, c < h, c < i, d < a, d < b, d < g, d < h, d < f, d < i, d < j, e < f, e < g, e < h, e < i, f < g, f < h, g < h, g < i. However d c and c d, e f and f e, etc. 268 CHAPTER 10. TYCHONOFF’S THEOREM 10.2.7 Deﬁnition. are said to be comparable if either x y or y x. Two elements x and y of a partially ordered set (X,
) 10.2.8 Remark. We saw in the order diagram above that the elements d and c are not comparable. Also 1 and e are not comparable. In N, Q, R, and Z with the usual orderings, every two elements are comparable. In Example 10.2.4, 3 and 5 are not comparable. 10.2.9 Deﬁnitions. A partially ordered set (X, ) is said to be linearly ordered (or totally ordered) if every two elements are comparable. The order is then said to be a linear order (or a total order.) The linear ordering is said to be a strict linear ordering (or a strict total ordering) if a b and b a =⇒ a = b, for a, b ∈ X. 10.2.10 Examples. The usual orders on R, Q, N, and Z are linear orders. The partial order of Example 10.2.4 is not a linear order (if U has at least two points). 10.2.11 Deﬁnition. s ∈ X is said to be the greatest element of X if x s, for all x ∈ X. Let (X, ) be a partially ordered set. Then an element Let (X, ) be a partially ordered set and Y a subset 10.2.12 Deﬁnition. of X. An element t ∈ X is said to be an upper bound for Y if y t, for all y ∈ Y . It is important to note that an upper bound for Y need not be in Y . 10.2. ZORN’S LEMMA 269 10.2.13 Deﬁnition. w ∈ X is said to be maximal if w x, with x ∈ X, implies w = x. Let (X, ) be a partially ordered set. Then an element 10.2.14 Remark. It is important to distinguish between maximal elements and greatest elements. Consider the order diagram in Example 10.2.6. There is no greatest element! However, j, h, i and f are all maximal elements. 10.2.15 Remark. We can now state Zorn’s Lemma. Despite the name “Lemma”, it is, in fact, an axiom and cannot be proved. It is equivalent to various other axioms of Set Theory such as the Axiom of Choice and the Well-Ordering [A partially ordered set (S, ) is said to be well-ordered if every nonTheorem. empty subset of S has a least element. The Well-Ordering Theorem states that there exists a well-ordering on every set. See, for example, Halmos [168] or Wilder [424].] For a discussion of Zorn’s Lemma, the Axiom of Choice and Tychonoﬀ ’s Theorem, see Remark A6.1.24. Also see Rubin and Rubin [348]. We shall take Zorn’s Lemma as one of the axioms of our set theory and so use it whenever we wish. 10.2.16 Axiom. Let (X, ) be a non-empty partially ordered set in which every subset which is linearly ordered has an upper bound. Then (X, ) has a maximal element. (Zorn’s Lemma) 10.2.17 Example. Let us apply Zorn’s Lemma to the lattice diagram of Example 10.2.6. There are many linearly ordered subsets: {i, g, b, a}, {g, b, a}, {b, a}, {g, b}, {i, g}, {a}, {b}, {g}, {i}, {i, b, a}, {i, g, a}, {i.a}, {g, a}, {h, g, e}, {h, e}, {g, e}, etc. Each of these has an upper bound – i, i, i, i, i, i, i, i, i, i, i, i, i, h, h, h, etc. Zorn’s In fact there are 4 maximal Lemma then says that there is a maximal element. elements, j, h, f and i. 270 CHAPTER 10. TYCHONOFF’S THEOREM Exercises 10.2 1. Let X = {a, b, c, d, e, f, u, v}. Draw the order diagram of the partially ordered set (X, ) where v < a, v > b, v < c, v < d, v < e, v < f, v < u, a < c, a < d, a < e, a < f, a < u, b < c, b > d, b < e, b < f, b < u, c < d, c < e, c < f, c < u, d < e, d < f, d < u, e < u, f < u. 2. In Example 10.2.3, state which of the following subsets of N is linearly ordered: (a) {21, 3, 7}; (b) {3, 6, 15}; (c) {2, 6, 12, 72}; (d) {1, 2, 3, 4, 5, ...}; (e) {5}. 3. Let (X, ) be a linearly ordered set. If x and y are maximal elements of X, prove that x = y. 4. Let (X, ) be a partially ordered set. If x and y are greatest elements of X, prove that x = y. 5. Let X = {2, 3, 4, 5, 6, 7, 8, 9, 10} be partially ordered as follows: x y if y is a multiple of x. Draw an order diagram and ﬁnd all the maximal elements of (X, ). Does (X, ) have a greatest element? 10.3. TYCHONOFF’S THEOREM 271 6.* Using Zorn’s Lemma 10.2.16 prove that every vector space V has a basis. [Hints: (i) Consider ﬁrst the case where V = {0}. (ii) Assume V = {0} and deﬁne B = {B : B is a set of linearly independent vectors of V.} Prove that B = Ø. (iii) Deﬁne a partial order on B by B1 B2 if B1 ⊆ B2. Let {Bi : i ∈ I} be any linearly ordered subset of B. Prove that A = i∈I Bi is a linearly independent set of vectors of V . (iv) Deduce that A ∈ B and so is an upper bound for {Bi : i ∈ I}. (v) Apply Zorn’s Lemma to show the existence of a maximal element of B. Prove that this maximal element is a basis for V .] 10.3 Tychonoﬀ ’s Theorem Let X be a set and F a family of subsets of X. Then 10.3.1 Deﬁnition. F is said to have the ﬁnite intersection property or (F.I.P.) if for any ﬁnite number F1, F2, . . . , Fn of members of F, F1 ∩ F2 ∩ · · · ∩ Fn = Ø. 272 CHAPTER 10. TYCHONOFF’S THEOREM Let (X, τ ) be a topological space. Then (X, τ ) is 10.3.2 Proposition. compact if and only if every family F of closed subsets of X with the ﬁnite intersection property satisﬁes F ∈F F = Ø. Assume that every family F of closed subsets of X with the ﬁnite Proof. intersection property satisﬁes F ∈F F = Ø. Let U be any open covering of X. Put F equal to the family of complements of members of U . So each F ∈ F is closed in (X, τ ). As U is an open covering of X, F ∈F F = Ø. By our assumption, then, F does not have the ﬁnite intersection property. So for some F1, F2, . . . , Fn in F , F1 ∩F2 ∩· · ·∩Fn = Ø. Thus U1 ∪U2 ∪· · ·∪Un = X, where Ui = X \Fi, i = 1, . . . , n. So U has a ﬁnite subcovering. Hence (X, τ ) is compact. The converse statement is proved similarly. Let X be a set and F a family of subsets of X with the 10.3.3 Lemma. ﬁnite intersection property. Then there is a maximal family of subsets of X that contains F and has the ﬁnite intersection property. Let Z be the collection of all families of subsets of X which contain F Proof. and have the ﬁnite intersection property. Deﬁne a partial order on Z as follows: if F1 and F2 are in Z then put F1 F2 if F1 ⊆ F2. Let Y be any linearly ordered subset of Z. To apply Zorn’s Lemma 10.2.16 we need to verify that Y has an upper bound. We claim that Y∈Y Y is an upper bound for Y . Clearly this contains F , so we have to show only that it has the ﬁnite intersection property. So let S1, S2, . . . , Sn ∈ Y∈Y Y. Then each Si ∈ Yi, for some Yi ∈ Y . As Y is linearly ordered, one of the Yi contains all of the others. Thus S1, S2, . . . , Sn all belong to that Yi. As Yi has the ﬁnite intersection property, S1 ∩ S2 ∩ · · · ∩ Sn = Ø. Hence Y∈Y Y has the ﬁnite intersection property and is, therefore, an upper bound in X of Y . Thus by Zorn’s Lemma, Z has a maximal element. We can now prove the much heralded Tychonoﬀ Theorem. 10.3. TYCHONOFF’S THEOREM 273 10.3.4 Theorem. any family of topological spaces. Then each (Xi, τ i) is compact. (Tychonoﬀ ’s Theorem) Let {(Xi, τ i) : i ∈ I} be i∈I (Xi, τ i) is compact if and only if Proof. We shall use Proposition 10.3.2 to show that (X, τ ) = i∈I (Xi, τ i) is compact, if each (Xi, τ i) is compact. Let F be any family of closed subsets of X with the ﬁnite intersection property. We have to prove that F ∈F F = Ø. By Lemma 10.3.3 there is a maximal family H of (not necessarily closed) subsets of (X, τ ) that contains F and has the ﬁnite intersection property. We shall prove that F ∈F F = Ø, since each F ∈ F is closed. H∈H H = Ø, from which follows the required result As H is maximal with the property that it contains F and has the ﬁnite intersection property, if H1, H2, . . . , Hn ∈ H, for any n ∈ N, then the set H = H1 ∩H2 ∩· · ·∩Hn ∈ H. Suppose this was not the case. Then the set H = H∪{H} would properly contain H and also have the property that it contains F and has the ﬁnite intersection property. This is a contradiction to H being maximal. So H = H and H = H1 ∩ H2 ∩ · · · ∩ Hn ∈ H. 1 ∩ H 1 ∩ H Let S be any subset of X that intersects non-trivially every member of H. We 2, . . . , H 1, H claim that H∪{S} has the ﬁnite intersection property. To see this let H m be members of H. We shall show that S ∩ H m = Ø. By the previous paragraph, H m) = Ø. Hence H ∪ {S} has the ﬁnite intersection property and contains F . Again using the fact that H is maximal with the property that it contains F and has the ﬁnite intersection property, we see that S ∈ H. So by assumption S ∩ (H Fix i ∈ I and let pi : i∈I (Xi, τ i) → (Xi, τ i) be the projection mapping. Then the family {pi(H) : H ∈ H} has the ﬁnite intersection property. Therefore the family {pi(H) : H ∈ H} has the ﬁnite intersection property. As (Xi, Ti) is compact, H∈H pi(H). Then for each i ∈ I, we can ﬁnd a point xi ∈ H∈H pi(H) = Ø. So let xi ∈ H∈H pi(H). Put x = i∈I xi ∈ X. We shall prove that x ∈ h∈H H. Let O be any open set containing x. Then (Ui), where Ui ∈ τ i, O contains a basic open set about x of the form i∈J p−1 i xi ∈ Ui and J is a ﬁnite subset of I. As xi ∈ pi(H), Ui ∩ pi(H) = Ø, for all H ∈ H. Thus p−1 (Ui) ∩ H = Ø, for all H ∈ H. By the observation above, this i implies that p−1 (Ui) ∈ H, for all i ∈ J . As H has the ﬁnite intersection property, (U1) ∩ H = Ø, for all H ∈ H. So O ∩ H = Ø for all H ∈ H. Hence x ∈ i I∈I (Xi, τ i) is compact, then by Propositions 7.2.1 and 10.1.5 i∈J p−1 i H∈H H, as required. (i) each (Xi, τ i) is compact. Conversely, if 274 CHAPTER 10. TYCHONOFF’S THEOREM 10.3.5 Notation. space (Ia, τ a) be homeomorphic to [0, 1]. Then the product space denoted by IA and referred to as a cube. Let A be any set and for each a ∈ A let the topological a∈A(Ia, τ a) is Observe that I N is just the Hilbert cube which we also denote by I∞. 10.3.6 Corollary. For any set A, the cube IA is compact. 10.3.7 Proposition. homeomorphic to a subspace of the cube IX . Let (X, d) be a metric space. Then it is Proof. Without loss of generality, assume d(a, b) 1 for all a and b in X. For each a ∈ X, let fa be the continuous mapping of (X, d) into [0, 1] given by fa(x) = d(x, a). That the family {fa : a ∈ X} separates points and closed sets is easily shown (cf. the proof of Theorem 9.4.11). Thus, by Corollary 10.1.10 of the Embedding Lemma, (X, d) is homeomorphic to a subspace of the cube IX . This leads us to ask: Which topolo
gical spaces are homeomorphic to subspaces of cubes? We now address this question. Let (X, τ ) be a topological space. Then (X, τ ) is 10.3.8 Deﬁnitions. said to be completely regular if for each x ∈ X and each open set U x, there exists a continuous function f : (X, τ ) −→ [0, 1] such that f (x) = 0 and If (X, τ ) is also Hausdorﬀ, then it is said to be f (y) = 1 for all y ∈ X \ U . -space). Tychonoﬀ space (or a T 3 1 2 10.3. TYCHONOFF’S THEOREM 275 10.3.9 Proposition. induced on X by d. Then (X, τ ) is a Tychonoﬀ space. Let (X, d) be a metric space and τ the topology Let a ∈ X and U be any open set containing a. Then U contains an Proof. open ball with centre a and radius ε, for some ε > 0. Deﬁne f : (X, d) −→ [0, 1] by f (x) = min 1, , d(x, a) ε for x ∈ X. Then f is continuous and satisﬁes f (a) = 0 and f (y) = 1, for all y ∈ X \ U . As (X, d) is also Hausdorﬀ, it is a Tychonoﬀ space. 10.3.10 Corollary. The space [0, 1] is a Tychonoﬀ space. 10.3.11 Proposition. regular spaces, then i∈I (Xi, τ i) is completely regular. If {(Xi, τ i) : i ∈ I} is any family of completely Let a = i∈I ai ∈ Proof. there exists a ﬁnite subset J of I and sets Ui ∈ τ i such that i∈I Xi and U be any open set containing a. Then a ∈ i∈I Ui ⊆ U where Ui = Xi for all i ∈ I \ J. As (Xj, Tj) is completely regular for each j ∈ J , there exists a continuous mapping fj : (Xj, τ j) −→ [0, 1] such that fj(aj) = 0 and i∈I (Xi, τ i) −→ [0, 1], where pj fj(y) = 1, for all y ∈ Xj \ Uj. Then fj ◦ pj : denotes the projection of i∈I (Xi, τ i) onto (Xj, τ j). If we put f (x) = max{fj ◦ pj(x) : j ∈ J}, for all x ∈ i∈I Xi, then f : i∈I (Xi, τ i) −→ [0, 1] is continuous (as J is ﬁnite). Further, f (a) = 0 while f (y) = 1 for all y ∈ X \ U . So i∈I (Xi, τ i) is completely regular. 276 CHAPTER 10. TYCHONOFF’S THEOREM The next proposition is easily proved and so its proof is left as an exercise. 10.3.12 Proposition. spaces, then i∈I (Xi, τ i) is Hausdoroﬀ. If {(Xi, τ i) : i ∈ I} is any family of Hausdorﬀ Proof. Exercise. 10.3.13 Corollary. then i∈I (Xi, τ i) is a Tychonoﬀ space. If {(Xi, τ i) : i ∈ I} is any family of Tychonoﬀ spaces, 10.3.14 Corollary. For any set X, the cube IX is a Tychonoﬀ space. 10.3.15 Proposition. Every subspace of a completely regular space is completely regular. Proof. Exercise. 10.3.16 Corollary. Every subspace of a Tychonoﬀ space is a Tychonoﬀ space. Proof. Exercise. 10.3. TYCHONOFF’S THEOREM 277 10.3.17 Proposition. If (X, τ ) is any Tychonoﬀ space, then it is homeomorphic to a subspace of a cube. Proof. Let F be the family of all continuous mappings f : (X, τ ) −→ [0, 1]. Then if follows easily from Corollary 10.1.10 of the Embedding Lemma and the deﬁnition of completely regular, that the evaluation map e : (X, τ ) → IF is an embedding. Thus we now have a characterization of the subspaces of cubes. Putting together Proposition 10.3.17 and Corollaries 10.3.14 and 10.3.16 we obtain: 10.3.18 Proposition. cube if and only if it is a Tychonoﬀ space. A topological space (X, τ ) can be embedded in a 10.3.19 Remark. We now proceed to show that the class of Tychonoﬀ spaces is quite large and, in particular, includes all compact Hausdorﬀ spaces. A topological space (X, τ ) is said to be a normal 10.3.20 Deﬁnitions. space if for each pair of disjoint closed sets A and B, there exist open sets U and V such that A ⊆ U , B ⊆ V and U ∩ V = Ø. A normal space which is also Hausdorﬀ is said to be a T4-space. 10.3.21 Remarks. In Exercises 6.1 #9 it is noted that every metrizable space is a normal space. A little later we shall verify that every compact Hausdorﬀ space is normal. First we shall prove that every normal Hausdorﬀ space is a Tychonoﬀ space (that is, every T4-space is a T -space). 3 1 2 Putting C = X \ B, and K = X \ V in Deﬁnition 10.3.20 of a normal space, we see that a topological space (X, τ ) is a normal space if and only if for every closed set A and open set C with A ⊆ C, there exists a closed set K with A ⊆ Int(K) ⊆ K ⊆ C. 278 CHAPTER 10. TYCHONOFF’S THEOREM (Urysohn’s Lemma) Let (X, τ ) be a topological 10.3.22 Theorem. space. Then (X, τ ) is normal if and only if for each pair of disjoint closed sets A and B in (X, τ ) there exists a continuous function f : (X, τ ) −→ [0, 1] such that f (a) = 0 for all a ∈ A, and f (b) = 1 for all b ∈ B. Proof. Then U = f −1([0, 1 B ⊆ V , and A ∩ B = Ø. Hence (X, τ ) is normal. Assume that for each A and B an f with the property stated above exists. 2, 1]) are open in (X, τ ) and satisfy A ⊆ U , 2)) and V = f −1((1 open subsets of X, where the set D is given by D = k Conversely, assume (X, τ ) is normal. We shall construct a family {Ui : i ∈ D} of 2n : k = 1, 2, . . . , 2n, n ∈ N. So D is a set of dyadic rational numbers, such that A ⊆ Ui, Ui ∩ B = Ø, and . As (X, τ ) is normal, for any pair A, B of disjoint closed such that A ⊆ U 1 . So we 2 ⊆ BC where the superscript C is used to denote complements d1 d2 implies Ud1 ⊆ Ud2 sets, there exist disjoint open sets U 1 2 and B ⊆ V 1 2 and V 1 2 and BC = X \ B). have A ⊆ U 1 2 in X (that is Now consider the disjoint closed sets A and U C 1 2 and U C and V 1 1 4 2 are disjoint closed sets there exists disjoint open sets U 3 4 disjoint open sets U 1 4 such that A ⊆ U 1 4 and B ⊆ V 3 4 . So we have . Again, by normality, there exist ⊆ V 1 4 and V 3 4 . Also as V C 1 2 such that V C 1 2 and Continuing by induction we obtain open sets Ud and Vd, for each d ∈ D, such ⊆ BC. that 2−n ⊆ U2.2−n ⊆ V C A ⊆ U2−n ⊆ V C So we have, in particular, that for d1 d2 in D, Ud1 ⊆ Ud2 Now we deﬁne f : (X, τ ) −→ [0, 1] by f (x) = 2.2−n ⊆ · · · ⊆ U(2n−1)2−n ⊆ V C . d∈D Ud d∈D Ud. Observe ﬁnally that since A ⊆ Ud, for all d ∈ D, f (a) = 0 for all a ∈ A. Also d∈D Ud and so f (b) = 1. So we have to show only that f is inf{d : x ∈ Ud}, 1, (2n−1)2−n ⊆ BC. if x ∈ if x /∈ if b ∈ B, then b /∈ continuous. 10.3. TYCHONOFF’S THEOREM 279 Let f (x) = y, where y = 0, 1 and set W = (y − ε, y + ε), for some ε > 0 (with ). As D is dense in [0, 1], we can choose d0 and d1 such that y − ε < d0 < y < d1 < y0 + ε. Then, by the deﬁnition of f , x ∈ U = Ud1 \ U d0 and the open set U satisﬁes f (u) ⊆ W . If y = 1 then we put W = (y − ε, 1], choose d0 . Again f (U ) ⊆ W . Finally, if y = 0 such that y − ε < d0 < 1, and set U = X \ U d0 then put W = [0, y + ε), choose d1 such that 0 < d1 < Y + ε and set U = Ud1 to again obtain f (U ) ⊆ W . Hence f is continuous. 10.3.23 Corollary. Tychonoﬀ space; that is, every T4-space is a T homeomorphic to a subspace of a cube. If (X, τ ) is a Hausdorﬀ normal space then it is a -space. Consequently it is 3 1 2 10.3.24 Proposition. Every compact Hausdorﬀ space (X, τ ) is normal. Let A and B be disjoint closed subsets of (X, τ ). Fix b ∈ B. Then, as Proof. (X, τ ) is Hausdorﬀ, for each a ∈ A, there exist open sets Ua and Va such that a ∈ Ua, b ∈ Va and Ua∩Va = Ø. So {Ua : a ∈ A} is an open covering of A. As A is compact, there exists a ﬁnite subcovering Ua1, Ua2, . . . , Uan. Put Ub = Ua1 ∪ Ua2 ∪ · · · ∪ Uan and Vb = Va1 ∩ Va2 ∩ . . . Van. Then we have A ⊆ Ub, b ∈ Vb, and Ub ∩ Vb = Ø. Now let b vary throughout B, so we obtain an open covering {Vb : b ∈ B} of B. As B is compact, there exists a ﬁnite subcovering Vb1, Vb2, . . . , Vbm . Set . Then A ⊆ U , B ⊆ V , and V = Vb1 ∪ Vb2 ∪ · · · ∪ Vbm U ∩ V = Ø. Hence (X, τ ) is normal. and U = Ub1 ∩ Ub2 ∩ · · · ∩ Ubm 10.3.25 Corollary. cube. Every compact Hausdorﬀ space can be embedded in a 280 CHAPTER 10. TYCHONOFF’S THEOREM 10.3.26 Remark. We can now prove the Urysohn metrization theorem which provides a suﬃcient condition for a topological space to be metrizable. It also provides a necessary and suﬃcient condition for a compact space to be metrizable – namely that it be Hausdorﬀ and second countable. A topological space (X, τ ) is said to be regular if 10.3.27 Deﬁnitions. for each x ∈ X and each U ∈ τ such that x ∈ U , there exists a V ∈ τ with x ∈ V ⊆ U . If (X, τ ) is also Hausdorﬀ it is said to be a T3-space. 10.3. TYCHONOFF’S THEOREM 281 10.3.28 Remark. So, from Corollary 10.3.23, every T4-space is a T3-space. hierarchy: It is readily veriﬁed that every T3 1 -space is a T3-space. 2 Indeed we now have a compact Hausdorﬀ ⇒ T4 ⇒ T 3 1 2 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0 282 CHAPTER 10. TYCHONOFF’S THEOREM metrizable ⇒ T4 ⇒ T 3 1 2 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0. 10.3. TYCHONOFF’S THEOREM 283 10.3.29 Proposition. (X, τ ) is metrizable. Every normal second countable Hausdorﬀ space It suﬃces to show that (X, τ ) can be embedded in the Hilbert cube Proof. I∞. By Corollary 9.4.10, to verify this it is enough to ﬁnd a countable family of continuous maps of (X, τ ) into [0, 1] which separates points and closed sets. Let B be a countable basis for τ , and consider the set S of all pairs (V, U ) such that U ∈ B, V ∈ B and V ⊆ U . Then S is countable. For each pair (V, U ) in S we can, by Urysohn’s Lemma 10.3.22, ﬁnd a continuous mapping fV U : (X, τ ) −→ [0, 1] such that fV U (V ) = 0 and fV U (X \ U ) = 1. Put F equal to the family of functions, f , so obtained. Then F is countable. To see that F separates points and closed sets, let x ∈ X and W any open set containing x. Then there exists a U ∈ B such that x ∈ U ⊆ W . By Remark 10.3.28, (X, τ ) is regular and so there exists a set P ∈ τ such that x ∈ P ⊆ P ⊆ U . Therefore these exists a V ∈ B with x ∈ V ⊆ P . So x ∈ V ⊆ P ⊆ U . Then (V, U ) ∈ S and if fV U is the corresponding mapping in F , then fV U (x) = 0 /∈ {1} = fV U (X \ W ). 284 CHAPTER 10. TYCHONOFF’S THEOREM 10.3.30 Lemma. Every regular second countable space (X, τ ) is normal. Let A and B be disjoint closed subsets of (X, T ) and B a countable basis Proof. for τ . As (X, τ ) is regular and X \ B is an open set, for each a ∈ A there exists a Va ∈ B such that V a ⊆ X \ B. As B is countable we can list the members {Va : a ∈ A} so obtained by Vi, i ∈ N; that is, A ⊆ ∞ i=1 Vi and V i ∩ B = Ø, for all i ∈ N. Similarly we can ﬁnd sets Ui in B, i ∈ N, such that B ⊆ ∞ i=1 Ui and U i ∩A = Ø, for all i ∈ N. Now deﬁne U 1 ∩ V So U 1 = U1 \ V 1 and V 1
= V1 \ U 1, U Then we inductively deﬁne 1 ∩ B = U1 ∩ B, U n = Un \ n V i and V n = Vn \ n U i and V 1 ∩ A = V1 ∩ A. i=1 n ∩ B = Un ∩ B, i=1 n ∩ A = An ∩ A. and V So that U n ∈ τ , U n=1 U n ∈ τ , V Now put U = ∞ Then U ∩ V = Ø, U ∈ τ , Hence (X, τ ) is a normal space. n and V = ∞ n=1 V n. V ∈ τ , A ⊆ V , and B ⊆ U . We can now deduce from Proposition 10.3.29. and Lemma 10.3.30 the Urysohn Metrization Theorem, which generalizes Proposition 10.3.29. 10.3.31 Theorem. second countable Hausdorﬀ space is metrizable. (Urysohn’s Metrization Theorem) Every regular From Urysohn’s Metrization Theorem, Proposition 9.4.4, and Proposition 9.4.17, we deduce the following characterization of metrizability for compact spaces. 10.3.32 Corollary. Hausdorﬀ and second countable. A compact space is metrizable if and only if it is 10.3. TYCHONOFF’S THEOREM 285 10.3.33 Remark. As mentioned in Remark 10.3.21, every metrizable space is normal. It then follows from Proposition 9.4.17 that every separable metric space is normal, Hausdorﬀ, and second countable. Thus Uryshohn’s Theorem 9.4.11, which says that every separable metric space is homeomorphic to a subspace of the Hilbert cube, is a consequence of (the proof of ) Proposition 10.3.29. 10.3.34 Remark. There are some surprises in store as regards products of separable spaces. You might reasonably expect that a ﬁnite product of separable spaces is separable. Indeed it would not be unexpected to hear that a countable product of separable spaces is separable eg Nℵ0 and Rℵ0 are separable. And these are all true. But it would be surprising to hear that Rc is separable. these follow from the Hewitt-MarczewskiAll Pondiczery Theorem (Hewitt [179]; Marczewski [276]; Pondiczery [329]) below, due to Edwin Hewitt (1920–1999), Edward Marczewski (1907– 1976) and E.S. Pondiczery.1 This theorem is not only surprising but also very informative, as we shall see, in the next section as it will tell us that the Stone-˘Cech compactiﬁcation of many spaces are in fact huge. Hewitt Marczewski 1E.S. Pondiczery was a pseudonym invented by Ralph P. Boas Jr, Frank Smithies and colleagues. 286 CHAPTER 10. TYCHONOFF’S THEOREM Let (D, τ d) be a discrete topological space, F the set 10.3.35 Lemma. of all ﬁnite subsets of D and for each F ∈ F , let τ F be the discrete topology on F so that (F, τ F ) is a subspace of (D, τ d). Let I be any index set and AF a dense subset of the product space (F, τ F )I , for each F ∈ F. If A = AF , F ∈F then A is a dense subset of (D, τ d)I . Let U be any open set in (D, τ d)I . Since (D, τ d) is discrete, the deﬁnition Proof. of the product topology, Deﬁnition 10.1.1, shows that there exists i1, i2, . . . , ik ∈ I and xi1, xi2, . . . , xik ∈ D such that U ⊇ {xi1} × {xi2} × . . . {xik} × DI\{i1,i2,...,,ik}. Put F = {xi1, xi2, . . . , xik}. As AF is dense in (F, τ F )I and {xi1} × {xi2} × . . . {xik} × F I\{i1,i2,...,ik} is an open set in (F, τ F )I , {xi1} × {xi2} × . . . {xik} × F I\{i1,i2,...,ik} ∩ AF = Ø. This implies {xi1} × {xi2} × . . . {xik} × DI\{i1,i2,...,ik} ∩ AF = Ø. This in turn imples that U ∩ AF = Ø, and so U ∩ A = Ø. Hence A is indeed dense in (D, τ d)I , as required. 10.3. TYCHONOFF’S THEOREM 287 10.3.36 Proposition. topological space (X, τ ) = {0, 1}2m space {0, 1}. Then (X, τ ) has a dense subspace of cardinality m. Let m be an inﬁnite cardinal number and let the be the product of 2m copies of the disrete Proof. card (P(M )) = 2m. Let M be a set with card M = m. The power set P(M ) has We are required to ﬁnd a dense subset Y of (X, τ ) of cardinality m. It suﬃces to ﬁnd a set S and a function φ : S → X with φ(S) a dense subset of (X, τ ) and card (S) = m. Let F(M ) be the set of all ﬁnite subsets of M . So card (F(M )) = m. (Exercise.) Let F(F(M )) be the set of all ﬁnite subsets of F(M ). Then card (F(F(M ))) = m. Put S = F(M ) × F(F(M )). So card (S) = m. We are now looking for a function φ of S into X. Recall that X = {0, 1}2m functions from P(M ) to {0, 1}. So for every subset T of S, φ(T ) is a function from P(M ) to {0, 1}. = {0, 1}P(M ), and {0, 1}P(M ) is the set of all Let N be a subset of M ; that is, N ∈ P(M ). Further, let F ∈ F(M ) and F ∈ F(F(M )). 1, 0, Deﬁne φ : S → X by φ(F, F)(N ) = if N ∩ F ∈ F if N ∩ F /∈ F. Let x ∈ X and let Nj, j ∈ J , be a ﬁnite number of distinct subsets of M . Put K = {(j1, j2) : j1, j2 ∈ J, j1 = j2} and for each (j1, j2) ∈ K, deﬁne Tj1j2 = (Nj1 ∪ Nj2) \ (Nj1 ∩ Nj2). Let σ : K → (j1,j2)∈K Tj1,j2 be any map such that σ(j1, j2) ∈ Tj1j2 , (j1, j2) ∈ K. Set F = {Nj ∩ σ(K) : j ∈ J and x(Nj) = 1}. [Recall that x ∈ {0, 1}P(M ), that is x is a mapping from P(M ) to {0, 1}.] Then it is readily veriﬁed that φ(σ(K), F)(Nj) = x(Nj), for all j ∈ J , which completes the proof.2 2The proof here of this proposition is based on that in Blair [46] which is closely related to that of Theorem 9.2 of Gillman and Jerison [156] that every inﬁnite set, X, has 22card (X) distinct ultraﬁlters. See also Proposition A6.4.11. 288 CHAPTER 10. TYCHONOFF’S THEOREM Let (X, τ ) be a topological space and let m be the 10.3.37 Deﬁnition. least cardinal number such that (X, τ ) has a dense subset of cardinality m. Then (X, τ ) is said to have density character m. Clearly a topological space is separable if and only if its density character is less than or equal to ℵ0. So we can restate Proposition 10.3.36 in the following way. 10.3.38 Proposition. (X, τ ) = {0, 1}2m has density character at most m. For any inﬁnite cardinal m, the topological space 10.3.39 Corollary. (F, τ F ), the product space (X, τ ) = (F, τ F )2m m. For any inﬁnite cardinal m and any ﬁnite discrete space has density character at most Let n be a positive integer such that 2n > card F and let f be any Proof. (continuous) map of the discrete space {0, 1}n onto the discrete space (F, τ F ). onto (F, τ F )2m Then there is a continuous map of ({0, 1}n)2m . As n is ﬁnite and m is an inﬁnite cardinal, it is easily checked that ({0, 1}n)2m is homeomorphic to . So there is a continuous map of {0, 1}2m {0, 1}2m . As the image under a continuous map of a dense subset is a dense subset of the image, it follows that the density character of (F, τ F )2m is less than or equal to the density character of {0, 1}2m , which by Proposition 10.3.38 is less than or equal to m, as required. onto (F, τ F )2m Let (D, τ d) be a discrete topological space of 10.3.40 Proposition. cardinality less than or equal to m, for m any inﬁnite cardinal number. Then (D, τ d)2m has density character less than or equal to m. Proof. This follows immediately from Lemma 10.3.35 and Corollary 10.3.39. 10.3. TYCHONOFF’S THEOREM 289 (Hewitt-Marczewski-Pondiczery Theorem) Let m 10.3.41 Theorem. be an inﬁnite cardinal number, I a set of cardinality less than or equal to 2m, and (Xi, τ i), i ∈ I, topological spaces each of density character less than or (Xi, τ i) is less equal to m. Then the density character of the product space i∈I than or equal to m. In particular, if (X, τ ) is any topological space of density character less than or equal to m, then the density character of the product space (X, τ )2m is less than or equal to m. By Proposition 10.3.40 each (Xi, τ i) has a dense subset which is a Proof. (Xi, τ i) continuous image of the discrete space (D, τ d) of cardinality m and so i∈I has a dense subset which is a continuous image of the product space (D, τ d)2m . As (D, τ d)2m has a dense subset of cardinality less than or equal to m, the product (Xi, τ i) also has a dense subset of cardinality less than or equal to m; that space i∈I is, it has density character less than or equal to m. 10.3.42 Corollary. is a separable space. In particular, Rc is separable. If (X, τ ) is a separable topological space, then (X, τ )c We shall conclude this section with extension theorems; the ﬁrst and most important is the Tietze Extension Theorem which is of interest in itself, but also is useful in our study of the Stone–˘Cech compactiﬁcation in the next section. We shall prove various special cases of the Tietze Extension Theorem before stating it in full generality. 290 CHAPTER 10. TYCHONOFF’S THEOREM 10.3.43 Proposition. Let (X, τ ) be a Hausdorﬀ topological space. The following conditions are equivalent: (i) (X, τ ) is normal; (ii) for every closed subspace (S, τ 1) of (X, τ ) and each continuous map φ of (S, τ 1) into the closed unit interval [0, 1] with the euclidean topology, there exists a continuous extension Φ : (X, τ ) → [0, 1] of φ. 3 Assume that (ii) is true. Let A and B be disjoint closed subsets of Proof. (X, τ ). Put S = A ∪ B and deﬁne φ : S → R by φ(x) = 0, for x ∈ A, and φ(x) = 1, for x ∈ B. Then clearly φ : (A ∪ B, τ 1) → R is continuous, where τ 1 is the subspace topology of A ∪ B from (X, τ ). By hypothesis, there exists a continuous map Φ : (X, τ ) → R which extends φ. Let U and V be disjoint open sets in R containing 0 and 1, respectively. Then Φ−1(U ) and Φ−1(V ) are disjoint open sets containing A and B, respectively. So (X, τ ) is indeed a normal topological space; that is, (i) is true. Now assume that (i) is true. Firstly let us consider the case that φ : (S, τ 1) → [0, 1]. Deﬁne Sr = {x ∈ S : φ(x) r}, for r ∈ Q, and Ts = X \ {x ∈ S : φ(x) s}, for s ∈ Q ∩ (0, 1). We deﬁne the index set P by P = {(r, s) : r, s ∈ Q with 0 r < s < 1}. For convenience write P = {(rn, sn) : n ∈ N}. For brevity we shall denote the interior, Int(Y ), in (X, τ ) of any subset Y of X by Y 0. Our proof will use deﬁnition by mathematical induction. Noting that Sr1 is a closed set, Ts1 is an open set, Sr1 ⊆ Ts1 and (X, τ ) is normal, Remark 10.3.21 shows that there exists a closed set H1 in (X, τ ) such that Sr1 ⊆ H0 1 ⊆ H1 ⊆ Ts1. Next, assume that closed sets Hk have been constructed for all k < n ∈ N such that Srk ⊆ H0 k ⊆ Hk ⊆ Tsk, for k < n, and Hj ⊆ H0 k, when j, k < n, rj < rk and sj < sk. (1) (2) Deﬁne J = {j : j < n, rj < rn, and sj < sn} rk, and sn < sk}. and K = {k : k < n, rn < 3The proof here is based on that of Mandelkern [275]. In the
literature there are alternative proofs using uniform continuity. 10.3. TYCHONOFF’S THEOREM 291 Noting the deﬁnitions of Sr and Ts and using (1) and (2), we can apply Remark , to show that there exists a 10.3.21, with A = Srn ∪ j∈J closed set Hn in (X, τ ) such that Hj and C = Tsn ∩ k∈K H0 k Srn ∪ j∈J Hj ⊆ H0 n ⊆ Hn ⊆ Tsn ∩ H0 k. k∈K From this equation (3) one can verify that Srk ⊆ H0 k ⊆ Hk ⊆ Tsk, for k < n + 1, and Hj ⊆ H0 k, when j, k < n + 1, rj < rk and sj < sk. (3) (1) (2) Equations (1), (2), (1’) and (2’) complete our inductive deﬁnition of the closed sets Hn. Now we write Hrs for Hn where r = rn and s = sn. So that we have closed sets Hrs, for (r, s) ∈ P , with Sr ⊆ H0 rs ⊆ Hrs ⊆ Ts , for (r, s) ∈ P, and Hrs ⊆ H0 tu , when r < t and s < u. Now deﬁne Xr as follows: Xr =    X, Ø, s>0, 1) Hrs, For (r, s) ∈ P , choose t such that r < t < s. Then by (6) and (5), Xr ⊆ Hrt ⊆ H0 ts ⊆ Hts ⊆ Hsu = Xs From (7) we deduce that Xr ⊆ X0 u>s s , for (r, s) ∈ P with r < s. From the deﬁnitions of Sr & Tr, and equations (4) and (6) we have: Sr ⊆ S ∩ Xr = S ∩ Hrs ⊆ S ∩ Ts = Sr , for r ∈ Q ∩ [0, 1). (4) (5) (6) (7) (8) (9) So by (8) and noting that all the set containments in (9) are actually equality, s>r s>r we have found closed subsets {Xr : r ∈ Q} of (X, τ ) with the properties: for r, s ∈ Q, r < s , Xr ⊆ X0 s , and Xr ∩ S = Sr. (10) Finally deﬁne Φ(x) = inf{r : x ∈ Xr}, x ∈ X. By (6), Φ : (X, τ ) → [0, 1], and since φ(x) = inf{r : x ∈ Sr}, we have Φ(x) = φ(x), for all x ∈ S; that is, Φ If a, b ∈ R with a < b, then it follows immediately from the is an extension of φ. deﬁnition of Φ that Φ−1((a, b)) = {X0 s \ Xr : r, s ∈ Q and a < r < s < b}, and so Φ is continuous; that is, Φ is a continuous extension of φ, as required. 292 CHAPTER 10. TYCHONOFF’S THEOREM 10.3.44 Proposition. Let (X, τ ) be a Hausdorﬀ topological space. The following conditions are equivalent: (i) (X, τ ) is normal; (ii) for every closed subspace (S, τ 1) of (X, τ ) and each continuous map φ of (S, τ 1) into the open unit interval (0, 1) with the euclidean topology, there exists a continuous extension Φ : (X, τ ) → (0, 1) of φ. Proof. That (ii) implies (i) is proved analagously to that in Proposition 10.3.43. So assume that (i) is true and that φ is a continuous map of (S, τ 1) into (0, 1). We want to ﬁnd a continuous map Γ : X → (0, 1) such that Γ(x) = φ(x), for all x ∈ S. By Proposition 10.3.43, there exists a continuous map Φ : (X, τ ) → [0, 1], such that Φ(x) = φ(x), for all x ∈ X. Let D = {x : x ∈ X, Φ(x) ∈ {0, 1}}. Then S and D are disjoint closed sets. As (X, τ ) is a normal space, by Urysohn’s Lemma 10.3.22, there exists a continuous map θ : (X, τ ) → [ 1 2 for all x ∈ D. So if we deﬁne Γ : (X, τ ) → (0, 1) by Γ(x) = Φ(x).θ(x).θ(x) + 1 − θ(x), we can easily verify that Γ is a continuous extension of φ, as required. 2, 1] such that θ(x) = 1 for all x ∈ S, and θ(x) = 1 Let (S, τ 1) be a subspace of the topological space 10.3.45 Lemma. (X, τ ) and (Y, τ 2) and (Z, τ 3) homeomorphic topological spaces. If every continuous map φ : (S, τ 1) → (Y, τ 2) has a continuous extension Φ : (X, τ ) → (Y, τ 2), then every continuous map γ : (S, τ 2) → (Z, τ 3) also has a continuous extension Γ : (X, τ ) → (Z, τ 3). Proof. Exercise. As an immediate consequence of Proposition 10.3.44 and Lemma 10.3.45 we have: 10.3. TYCHONOFF’S THEOREM 293 10.3.46 Proposition. Let (X, τ ) be a Hausdorﬀ topological space. The following conditions are equivalent: (i) (X, τ ) is normal; (ii) for every closed subspace (S, τ 1) of (X, τ ) and each continuous map φ of (S, τ 1) → R, there exists a continuous extension Φ : (X, τ ) → R of φ. Let (Y, τ 1) be a subspace of a topological space 10.3.47 Deﬁnition. (X, τ ). Then (Y, τ 1) is said to be a retract of (X, τ ) if there exists a continuous map θ : (X, τ ) → (Y, τ 1) with the property that θ(y) = y, for all y ∈ Y . 10.3.48 Example. [0, 1] is a retract of R. (Verify this.) Let (S, τ 1) be a subspace of the topological space 10.3.49 Lemma. (X, τ ). Further, let (Y, τ 2) and (Z, τ 3) be topological spaces such that (Z, τ 3) is a retract of (Y, τ 2). If every continuous map φ : (S, τ 1) → (Y, τ 2) has a continuous extension Φ : (X, τ ) → (Y, τ 2), then every continuous map γ : (S, τ 1) → (Z, τ 3) also has a continuous extension Γ : (X, τ ) → (Z, τ 3). Let γ : (S, τ 1) → (Z, τ 3) be any continuous map. As (Z, τ 3) is a retract Proof. of (Y, τ 2), there exists a continuous map θ : (Y, τ 2) → (Z, τ 3) with θ(z) = z, for all z ∈ Z. As γ is also a continuous map of (S, τ 1) into (Y, τ 2), by assumption there exists a continuous extension Φ : (X, τ ) → (Y, τ 2) of γ. Putting Γ = θ ◦ Φ, we have that Γ : (X, τ ) → (Z, τ 3) is a continuous extension of γ : (S, τ 1) → (Z, τ 3), as required. 294 CHAPTER 10. TYCHONOFF’S THEOREM Let (S, τ 1) be a subspace of the topological space 10.3.50 Lemma. (X, τ ). Let I be any index set and (Y, τ i), i ∈ I, a set of topological spaces. If every continuous mapping φi : (S.τ 1) → (Yi, τ i), i ∈ I, has a continuous extension Φi : (X, τ ) → (Yi, τ i), then the product map φ : (S, τ 1) → i∈I φ(x) = i∈I φi(x), x ∈ S, has a continuous extension Φ : (X, τ ) → i∈I (Yi, τ i), given by (Yi, τ i). Proof. the result. Deﬁning Φ to be the product map of the Φi, i ∈ I, immediately yields Finally, using Lemmas 10.3.50 and 10.3.49, Proposition 10.3.46 and Example 10.3.48, we obtain (a rather general version of ) the Tietze Extension Theorem. (Tietze Extension Theorem) Let (X, τ ) be a 10.3.51 Theorem. Hausdorﬀ topological space, m any cardinal number, and (Y, τ 2) any inﬁnite topological space which is a retract of the product space Rm. The following conditions are equivalent: (i) (X, τ ) is normal; (ii) for every closed subspace (S, τ 1) of (X, τ ) and each continuous map φ of (S, τ 1) into (Y, τ 2), there exists a continuous extension Φ : (X, τ ) → (Y, τ 2) of the map φ. In particular, this is the case when (Y, τ 2) is any non-trivial interval in R. 10.3. TYCHONOFF’S THEOREM 295 The condition in Theorem 10.3.51(ii) that (S, τ 1) is closed 10.3.52 Remark. in (X, τ ) is necessary. For example let φ : (0, 1] → R be the map φ(x) = sin 1 x , for x ∈ (0, 1]. Then φ is continuous, but there is no extension of φ to a map Φ : [0, 1] → R which is continuous. (Verify this.) We end this section with a useful result on extending continuous maps from a dense subspace which shall be useful in our discussion of the Wallman compactiﬁcation in §A6.4 of Appendix 6 296 CHAPTER 10. TYCHONOFF’S THEOREM Let (S, τ 1) be a dense subspace of a topological 10.3.53 Proposition. space (X, τ ) and φ : (S, τ 1) → (K, τ 2) a continuous map of (S, τ 1) into a compact Hausdorﬀ space (K, τ 2). The following conditions are equivalent: (i) the map φ has a continuous extension Φ : (X, τ ) → (K, τ 2); (ii) for every pair of closed subsets C1, C2 of (K, τ 2), the inverse images φ−1(C1) and φ−1(C2) have disjoint closures in (X, τ ). Firstly assume (i) is true, that is the continuous extension Φ exists. By Proof. continuity, Φ−1(C1) and Φ−1(C2) are closed disjoint sets in (X, τ ). So Φ−1(C1) ∩ Φ−1(C2) = Φ−1(C1) ∩ Φ−1(C2) = Ø. Hence (ii) is true. Now assume that (ii) is true. For every x ∈ X, let N (x) be the set of all open neighbourhoods of x ∈ (X, τ ). Let F(x) = {φ(S ∩ N ) : N ∈ N (x)} (1) Each member of F(x) is obviously a closed subset of (K, τ 2). We shall verify that F(x) has the ﬁnite interesection property, for each x ∈ X. Let N1, N2, . . . , Nn ∈ N (x). Then φ(S ∩ N1) ∩ φ(S ∩ N2) ∩ · · · ∩ φ(S ∩ Nn) ⊇ φ(S ∩ N1 ∩ N2 ∩ · · · ∩ Nn) (2) As S is dense in (X, τ ), it intersects the open set N1 ∩ N2 ∩ · · · ∩ Nn non-trivially. So S ∩ N1 ∩ N2 ∩ · · · ∩ Nn = Ø which implies φ(S ∩ N1 ∩ N2 ∩ · · · ∩ Nn) = Ø. Thus F(x) has the ﬁnite intersection property. As (K, τ 2) is compact, Proposition 10.3.2 implies that Fi∈F(x) Fi = Ø, for each x ∈ X. Deﬁne Φ(x) = Fi, for each x ∈ X. (3) Fi∈F(x) We need to verify that for each x ∈ X, Φ(x) is a single point, and that Φ : (X, τ ) → (K, τ 2) is continuous. If Φ(x) is a single point, then, by the previous paragraph, Φ(x) = φ(x), for all x ∈ S. Suppose that y1, y2 ∈ Φ(x), for some x ∈ X with y1 = y2. 10.3. TYCHONOFF’S THEOREM 297 As (K, τ 2) is compact Hausdorﬀ, by Remark 10.3.28 it is regular and Hausdorﬀ and so there exist open neighbourhoods U1, U2 of y1, y2, respectively such that U1∩U2 = Ø. By our assumption, φ−1(U1)∩φ−1(U2) = Ø. Putting O1 = X\φ−1(U1) and O2 = X \ φ−1(U2), we have X = O1 ∪ O2. So x ∈ Oj for j = 1 or j = 2. Since Uj∩φ(S\φ−1(Uj) = Ø and Uj is an open set in (K, τ ), we have Uj∩φ(S \ φ−1(Uj) = Ø, which implies that yj /∈ φ(S \ φ−1(Uj)) = φ(S ∩ Oj) ∈ F(x). By (3), then, yj /∈ Φ(x). This is a contradiction and so our supposition was false, and Φ(x) is a single point for each x ∈ X. Our ﬁnal task is to show that Φ : (X, τ ) → (K, τ 2) is continuous. Let U be an open neighbourhood of Φ(x) in (K, τ 2). By (1) and (3), {Φ(x)} = φ(S ∩ N ) ⊆ U N ∈N (x) (4). This implies that (K \ φ(S ∩ N ) ⊇ K \ U . As K \ U is compact and each N ∈N (x) K \ φ(S ∩ N ) is open, there exist N1, N2, . . . , Nk ∈ N (x) such that (K \ φ(S ∩ N1)) ∪ (K \ φ(S ∩ N2)) ∪ · · · ∪ (K \ φ(S ∩ Nk)) ⊇ K \ U. So As k i=1 k i=1 φ(S ∩ Ni) ⊆ U (5). Ni = N ∈ N (x), (1), (2), (4) and (5) imply that Φ(z) ∈ φ(S ∩ N ) ⊆ U , for every z ∈ N ; that is, Φ(N ) ⊆ U . So Φ is indeed continuous. 298 CHAPTER 10. TYCHONOFF’S THEOREM Exercises 10.3 Lindelöf Spaces 1. A topological space (X, τ ) is said to be a Lindelöf space if every open covering of X has a countable subcovering. Prove the following statements. (i) Every regular Lindelöf space is normal. [Hint: use a method like that in Lemma 10.3.30. Note that we saw in Exercises 9.4 #8 that every second countable space is Lindelöf.] (ii) The Sorgenfrey line (R, τ 1) is a Lindelöf space. (iii) If (X, τ ) is a topological space which has a closed uncountable discrete subspace, then (X, τ ) is not a Lindelöf space. (iv) It follows from (iii) above and Exercises 8.1 #12 that the product space (R, τ 1) × (R, τ 1) is no
t a Lindelöf space. [Now we know from (ii) and (iv) that a product of two Lindelöf spaces is not necessarily a Lindelöf space.] (v) Verify that a topological space is compact if and only if it is a countably compact Lindelöf space. (See Exercises 7.2 #17.) 2. Prove that any product of regular spaces is a regular space. 3. Verify that any closed subspace of a normal space is a normal space. 4. If (X, τ ) is an inﬁnite connected Tychonoﬀ space, prove that X is uncountable. 10.3. TYCHONOFF’S THEOREM 299 kω-spaces 5. A Hausdorﬀ space (X, τ ) is said to be a kω-space if there is a countable collection Xn, n ∈ N of compact subsets of X, such that (a) Xn ⊆ Xn+1, for all n, (b) X = ∞ (c) any subset A of X is closed if and only if A ∩ Xn is compact for each n ∈ N. n=1 Xn, Prove that (i) every compact Hausdorﬀ space is a kω-space; (ii) every countable discrete space is a kω-space; (iii) R and R2 are kω-spaces; (iv) every kω-space is a normal space; (v) every metrizable kω-space is separable; (vi) every metrizable kω-space can be embedded in the Hilbert cube; (vii) every closed subspace of a kω-space is a kω-space; (viii) if (X, τ ) and (Y, τ ) are kω-spaces then (X, τ ) × (Y, τ ) is a kω-space. (ix) if S is an inﬁnite subset of the kω-space (X, τ ), such that S is not contained in any Xn, n ∈ N, then S has an inﬁnite discrete closed subspace; (x) if K is a compact subspace of the kω-space (X, τ ), then K ⊆ Xn, for some n ∈ N. (xi)* a topological space (X, τ ) is said to be σ-metrizable if X = ∞ n=1 Xn, where Xn ⊆ Xn+1 for each n ∈ N, and each Xn with its induced topology is a closed metrizable subspace of (X, τ ). If every convergent sequence in the σ-metrizable space (X, τ ) is contained in Xn, for some n ∈ N, then (X, τ ) is said to be strongly σ-metrizable. (α) every closed subspace of a σ-metrizable space is σ-metrizable; (β) every closed subspace of a strongly σ-metrizable space is a strongly σ-metrizable space; (γ) If (X, τ ) is strongly σ-metrizable, then every closed compact subspace K of (X, τ ) is contained in Xn, for some n ∈ N.4 6. Prove that every T -space is a T3-space. 3 1 2 7. Prove that for metrizable spaces the conditions (i) Lindelöf space, (ii) separable, and (iii) second countable, are equivalent. 4This result is in fact true without the assumption that K is a closed subspace, see Banakh [27]. 300 CHAPTER 10. TYCHONOFF’S THEOREM First Axiom of Countability 8. A topological space (X, τ ) is said to satisfy the ﬁrst axiom of countability (or to be ﬁrst countable if for each x ∈ X, there exists a countable family Ui, i ∈ N of open sets containing x, such that if V ∈ τ and x ∈ V , then V ⊇ Un for some n. (i) Prove that every metrizable space is ﬁrst countable. (ii) Verify that every second countable space is ﬁrst countable, but that the converse is false. [Hint: Consider discrete spaces.] (iii) If {(Xi, τ i) : i ∈ N}, is a countable family of ﬁrst countable spaces, prove that ∞ i=1(Xi, τ i) is ﬁrst countable. (iv) Verify that every subspace of a ﬁrst countable space is ﬁrst countable. (v) Let X be any uncountable set. Prove that the cube IX is not ﬁrst countable, and hence is not metrizable. [Note that IX is an example of a [compact Hausdorﬀ and hence] normal space which is not metrizable.] (vi) Generalize (v) above to show that if J is any uncountable set and each j∈J (Xj, τ j) (X, τ j) is a topological space with more than one point, then is not metrizable. 9. Prove that the class of all Tychonoﬀ spaces is the smallest class of topological spaces that contains [0, 1] and is closed under the formation of subspaces and cartesian products. 10. Prove that any subspace of a completely regular space is a completely regular space. 11. Using Proposition 8.6.8, prove that if (G, τ ) is a topological group, then (G, τ ) is a regular space. [It is indeed true that every topological group is a completely regular space, but this is much harder to prove.] 10.3. TYCHONOFF’S THEOREM 301 12. Prove that if {(Xi, τ i) : i ∈ I} is any set of connected spaces, then j∈I(Xi, τ i) is connected. i∈I xi ∈ i∈I Xi which diﬀer from x = [Hint: Let x = in coordinates. Prove that CX (x) ⊇ S. Then show that S is dense in Finally use the fact that CX (x) is a closed set.] i∈I Xi. Let S consists of the set of all points in at most a ﬁnite number of i∈I (Xi, τ i). i∈I xi 13. Let {(Xj, τ j) : j ∈ J} be any set of topological spaces. Prove that j∈j(Xj, τ j) is locally connected if and only if each (Xj, τ j) is locally connected and all but a ﬁnite number of (Xj, τ j) are also connected. 14. Let (R, τ 1) be the Sorgenfrey line. Prove the following statements. (i) (R, τ 1) is a normal space. (ii) If (X, τ ) is a separable Hausdorﬀ space, then there are at most c distinct continuous functions f : (X, τ ) → [0, 1]. (iii) If (X, τ ) is a normal space which has an uncountable closed discrete subspace, then there are at least 2c distinct continuous functions f : (X, τ ) → [0, 1]. [Hint: Use Urysohn’s Lemma.] (iv) Deduce from (ii) and (iii) above and Exercises 8.1 #12, that (R, τ 1)×(R, τ 1) is not a normal space. [We now know that the product of two normal spaces is not necessarily a normal space.] (v) A topological space (X, τ ) is said to be hereditarily separable if (X, τ ) and each of its subspaces are separable. Show that the Sorgenfrey line (R, τ 1) is hereditarily separable. (vi) Show that if (R, τ 1) is the Sorgenfrey line, then the product space (R, τ 1) × (R, τ 1), known as the Sorgenfrey plane is separable but not hereditarily separable. [So we see that the product of two heritarily separable spaces need not be hereditarily separable and that a subspace of a separable space need not be separable. [Hint: Show that the subspace {(x, −x) : x ∈ R} of the Sorgenfrey plane is an uncountable discrete space.] 302 CHAPTER 10. TYCHONOFF’S THEOREM 15. If S is a set of inﬁnite cardinality m, verify that the cardinality of the set of all ﬁnite subsets of S is also m. 16. Verify (1’) in the proof of Proposition 10.3.43. 17. Prove Lemma 10.3.45. 18. Prove that every closed interval (Y, τ ) of R is a retract of R. 19. Let (Y, τ 1) be a subspace of a topological space (X, τ ). Prove that (Y, τ 1) is a retract of (X, τ ) if and only for every topological space (Z, τ 2) and every continuous map φ : (Y, τ 1) → (Z, τ 2) can be extended to a continuous map Φ : (X, τ ) → (Z, τ 2). 20. Verify the statement in Remark 10.3.52. 10.3. TYCHONOFF’S THEOREM 303 k-spaces 21. A topological space (X, τ ) is said to be a k-space (or a compactly-generated space) if for each subset S of (X, τ ) and each compact subspace (K, τ 1) of (X, τ ), S is closed in (X, τ ) if and only if S ∩ K is closed in (K, τ 1). (i) Prove that a subset U of a k-space is open in (X, τ ) if and only if U ∩ K is an open subset of (K, τ 1). (ii) Prove that every compact Hausdorﬀ space, every metrizable space, every kω-space, and every Hausdorﬀ sequential space is a k-space. (iii) Is a closed subspace of a k-space necessarily a k-space? (iv) Is an open subspace of a k-space necessarily a k-space? σ-compact Spaces 22. A topological space (X, τ ) is said to be σ-compact if there exist compact subsets Kn, n ∈ N, of X such that X = n∈N Kn. Prove the following: (i) If X is any countable set, then for any topology τ on X, (X, τ ) is a σ- compact space. (ii) Every compact space (X, τ ) is σ-compact. (iii) For each n ∈ N, the euclidean space Rn is σ-compact. (iv) Every kω-space is σ-compact. Further, Q is an example of a σ-compact space which is not a kω-space. (v) The topological space I of all irrational numbers with the euclidean topology is not a σ-compact space. (See Exercise 5 above.) (vi) Every σ-compact space is a a Lindelöf space. (vii) Let (X, τ ) and (Y, τ 1) be σ-compact spaces. The product space (X, τ ) × (Y, τ 1) is σ-compact. Deduce for this that any ﬁnite product of σ-compact spaces is σ-compact. (viii)* Prove that if each of the topological spaces (Xn, τ n), n ∈ N, is homeomorphic (Xn, τ n) is not σ-compact and so an to the discrete space Z, then n∈N inﬁnite product (even a countably inﬁnite product) of σ-compact spaces is not necessarily σ-compact. 304 CHAPTER 10. TYCHONOFF’S THEOREM Hemicompact Spaces 23. A topological space (X, τ ) is said to be hemicompact if it has a sequence of compact subsets Kn, n ∈ N, such that every compact subset C of (X, τ ) satisﬁes C ⊆ Kn, for some n ∈ N. Prove the following: (i) In the above deﬁnition, X = n∈N (Deduce that every hemicompact Kn. space is σ-compact.) [Hint: Use the fact that each singleton set in (X, τ ) is compact.] (ii) R is hemicompact. (iii) Every kω-space is hemicompact. (Deduce that R is hemicompact.) (iv) If X is an uncountable set and τ is the discrete topology on X, then (X, τ ) is not hemicompact. (Deduce that a metrizable locally compact space is not necessarily hemicompact.) (v) A ﬁrst countable hemicompact space is locally compact. (Deduce that a metrizable hemicompact space is locally compact. Deduce that every ﬁrst countable kω space is locally compact) (vi) A locally compact Hausdorﬀ σ-compact space is hemicompact. (vii) The topological space Q of all rational numbers with its usual topology is σ-compact but not hemicompact. (viii) A hemicompact Hausdorﬀ k-space is a kω-space. (ix) If (X, τ ) is a hemicompact space and (Y, d) is a metric space, then the space, C(X, Y ), of all continuous functions f : (X, τ ) → (Y, d) with the compact-open topology is metrizable. (See Deﬁnitions A5.6.4(b).) [Hint: Let Kn, n ∈ N be compact subsets of X such that each compact set is a subset of some Kn, n ∈ N. For each n ∈ N and each f, g ∈ C(X, Y ) deﬁne dn(f, g) = sup x∈Kn d(f (x), g(x)). Put d(f, g) = ∞ n=1 1 2n dn(f, g) 1 + dn(f, g) and verify that this is a metric on C(X, Y ) and induces the compact-open topology.] 10.3. TYCHONOFF’S THEOREM 305 24. Using previous exercises, prove the correctness of the picture below: 306 CHAPTER 10. TYCHONOFF’S THEOREM Reﬁnement of a Cover 25. Let X be a set and for some index set I, let {Ai : i ∈ I} be a c
over of X; that Ai = X. Then {Bj : j ∈ J}, J some index set, is said is each Ai ⊆ X and i∈I to be a reﬁnement of the cover {Ai : i ∈ I} if {Bj : j ∈ J} is a cover of X and for each j ∈ J , there exists an i ∈ I such that Bj ⊆ Ai. (i) Prove that every cover {Ai : i ∈ I} of X is is also a reﬁnement of itself. (ii) Prove that every subcover of {Ai : i ∈ I} is a reﬁnement of {Ai : i ∈ I}. 26. A set S of subsets of a topological space (X, τ ) is said to be locally ﬁnite in (X, τ ) if each point x in (X, τ ) has a neighbourhood Nx such that Nx ∩ S = Ø, for all but a ﬁnite number of S ∈ S. Prove the following statements: (i) If the set S of subsets of any topological space (X, τ ) is ﬁnite, then S is locally ﬁnite. (ii) If the set S is such that every point of X lies in at most one S in S, then S is locally ﬁnite in (X, τ ) for any topology τ on X. (iii) Let X be an inﬁnite set and τ the ﬁnite-closed topology on X. If S is the set of all open sets in (X, τ ), then S is not locally ﬁnite in (X, τ ). (iv) Let S be a locally ﬁnite set of subsets of (X, τ ). Deﬁne T to be the set of all closed sets T = S for S ∈ S. Then T is locally ﬁnite. (v) If S is an inﬁnite set of subsets of the inﬁnite set X and (X, τ ) is a compact space, then S is not locally ﬁnite in (X, τ ). [Hint: Suppose that (X, τ ) is locally ﬁnite and for each x ∈ X, choose a neighbourhood Ux which intersects only ﬁnitely-many members of S nontrivially. Then {Ux : x ∈ X} is an open covering of the set X and by the compactness of X, this has a ﬁnite subcovering of X. Show this leads to a contradiction.] (vi) Let S be an uncountable set of subsets of X. If S is a cover of the space (X, τ ) and (X, τ ) is either a Lindelöf space or a second countable space, then S is not locally ﬁnite. 10.3. TYCHONOFF’S THEOREM 307 Shrinking Lemma for Normal Spaces 27. For an index set I, let {Ui : i ∈ I} be an open covering of the topological space (X, τ ). The open covering {Vi : i ∈ I} of (X, τ ) is said to be a shrinking of the cover {Ui : i ∈ I} if Vi ⊆ Vi ⊆ Ui, for every i ∈ I. Prove the following results leading to the Shrinking Lemma for Normal Spaces. Let (X, τ ) be a normal Hausdorﬀ space. (i) For each x ∈ X and every open neighbourhood U of x, there exists an open neighbourhood V of x such that x ∈ V ⊆ V ⊆ U . (ii) If V is a closed set and U is an open set such that V ⊆ U , then there exists an open set W such that V ⊆ W ⊆ W ⊆ U . (This property characterizes normality.) (iii) Let {U, V } be an open covering of X. Then there exists an open set W such that W ⊆ W ⊆ U and {W, V } is an open covering of X. (iv) Every ﬁnite open covering of (X, τ ) has a shrinking; that is, for every open covering {Ui : i = 1, 2, . . . , n} of X, there exists an open covering {Vi : i = 1, 2, . . . , n} of X, such that Vi ⊆ Vi ⊆ Ui, for i = 1, 2, . . . , n. (This property characterizes normality.) (v)* For every locally ﬁnite open covering {Ui : i ∈ N} of X, there exists an open covering {Vi : i ∈ N} of X, such that Vi ⊆ Vi ⊆ Ui, for i ∈ N. (Without the assumption that the open covering {Ui : i ∈ N} is locally ﬁnite, the Axiom of Choice can be used to prove this result would be false. This is related to Dowker spaces, named after Cliﬀord Hugh Dowker (1912–1982). A Dowker space (X, τ ) is a normal space for which the product space (X, τ ) × [0, 1] is not a normal space. See Rudin [350].) Dowker (vi)* (Shrinking Lemma for Normal Spaces) For I any index set and {Ui : i ∈ I} a locally ﬁnite open covering of X, there exists an open covering {Vi : i ∈ I} of X, such that Vi ⊆ Vi ⊆ Ui, for i ∈ I. (C. H. Dowker proved that if (X, τ ) satisﬁes the Shrinking Lemma for every open covering, rather than every locally ﬁnite open covering, then it is a paracompact space - a property introduced in Exercise 29 below.) [Hint: For the inﬁnite case you may use the Axiom of Choice in the form of the Well-Ordering Theorem, Transﬁnite Induction and/or Zorn’s Lemma.] 308 CHAPTER 10. TYCHONOFF’S THEOREM Partitions of Unity and Bernstein Polynomials 28. Let (X, τ ) be a topological space. A set {fi : i ∈ I} of continuous functions from (X, τ ) to the closed unit interval [0,1], for some index set I, is said to be a partition of unity if for every x ∈ X: (a) there is a neighbourhood Nx of x, such that for all but a ﬁnite number of i ∈ I, fi vanishes on Nx, that is fi(y) = 0, for all y ∈ Nx; and (b) i∈I fi(x) = 1. A partition of unity is said to be subordinate to a covering U , of X if each fi, i ∈ I, vanishes outside some U ∈ U .. (i) Verify that for each locally ﬁnite open cover U = {Ui : i ∈ I}, for an index set I, of a normal Hausdorﬀ space, there is a partition of unity which is subordinate to U . [Hint: Using Exercise 27 (vi) above, deduce that there is an open covering V = {Vi : i ∈ I} of (X, τ ) such that each Vi ⊆ Ui. Then using Urysohn’s Lemma 10.3.22, construct continuous functions gi : (X, τ ) → [0, 1], i ∈ I, such that gi(Vi) = {1}, gi(X \ Ui) = {0}. Finally deﬁne fi = gi i∈I gi , for each i ∈ I. Verify that the fi are properly deﬁned and have the required properties.] (ii) [Bernstein (Basis) Polynomials5] The Bernstein (basis) polynomials xi(1 − x)n−i, for i = of degree n ∈ N are deﬁned to be Bi,n(x) = n i 0, 1, . . . , n, Bi,n = 0, for i < n or i > n. Calculate the Bernstein basis polynomials of degree 1, 2, and 3. (There are two of degree 1, three of degree 2, and four of degree 3.) (iii) Verify that each Bernstein basis polynomial of degree n can be written in terms of Bernstein basis polynomials of degree n − 1; more explicitly, Bi,n(x) = (1 − x)Bi,n−1(x) + xBi−1,n−1(x). (iv) Using (iii) above, prove by mathematical induction that all Bernstein basis polynomials satisfy Bi,n(x) 0 when x ∈ [0, 1] and Bi,n(x) > 0 when x ∈ (0, 1). 5For a discussion of Bernstein polynomials in the context of the Weierstrass Approximation Theorem, see Remark A7.1.6. 10.3. TYCHONOFF’S THEOREM 309 (v) Verify that k i=0 Bi,k(x) = k−1 i=0 Bi,k−1(x) , where k ∈ N . (vi) Verify that the n + 1 Bernstein basis polynomials of degree n are a partition of unity by using (v) to show that n i=0 Bi,n(x) = n−1 i=0 Bi,n−1(x) = n−2 i=0 Bi,n−2(x) = · · · = 1 i=0 Bi,1(x) = (1−x)+x = 1. (vii) Noting that every polynomial can be written as a linear sum of {1, x, x2, . . . , xn}, prove that every polynomial of degree n can also be written as a linear sum of Bernstein basis polynomials of degree 1, 2, . . . , n. [Hint: Using mathematical induction, verify that xk = n−1 i=k−1 i k Bi,n(x).] n k (viii) Verify that the Bernstein basis polynomials, B0,n(x), B1,n(x), . . . , Bn,n(x) are linearly independent by showing that 0 = λ0B0,n(x)+λ1B1,n(x)+· · ·+λnBn,n(x) = 0 , for all x , =⇒ λi = 0 , for i = 1, 2, . . . , n. (ix) Some writers on this topic (foolishly) deﬁne a Bernstein polynomial Bn(x) xm(1 − x)n−m of Bernstein basis am to be any linear combination polynomials of degrees 1, 2, . . . , n, where a1, a2, . . . , an ∈ R. Using (vii) above, verify that with this deﬁnition every polynomial is a Bernstein n m=0 m n polynomial. Paracompactness 29. A topological space (X, τ ) is said to be paracompact6 if every open cover has an open reﬁnement that is locally ﬁnite7. Prove the following statements: (i) Every compact space is paracompact. (ii) Every closed subspace of a paracompact space is paracompact. 6Some authors include Hausdorﬀness in the deﬁnition of paracompact. 7Paracompact spaces were introduced into the literature by Jean Alexandre Eugène Dieudonné who also proved that every Hausdorﬀ paracompact space is a normal space. 310 CHAPTER 10. TYCHONOFF’S THEOREM (iii)** Every Fσ-set S in a regular paracompact space is paracompact. (iv) Every regular Lindelöf space is paracompact. (v) Rn is paracompact, for every n ∈ N. (vi) The Sorgenfrey line is paracompact. (vii) Let (X, τ ) be the product of uncountable many copies of Z. Prove that (X, τ ) is not paracompact. (viii) Deduce from (vii) above, that if (X, τ ) is the product of uncountable many copies of any inﬁnite discete space, then (X, τ ) is not paracompact. (ix) Deduce from (viii) above, that if (X, τ ) is the product of uncountable many copies of R, then (X, τ ) is not paracompact. (x) Would (ix) above remain true, if R were replaced by an arbitrary inﬁnite non-compact metrizable space? (xi) Every Hausdorﬀ Lindelöf space is paracompact. (xii)* Every Hausdorﬀ paracompact space is a regular space. (xiii)* Every Hausdorﬀ paracompact space is a T4-space; that is a Hausdorﬀ normal space. (xiv) The space (X, τ ) is a paracompact Hausorﬀ space if and only if for every open cover U of (X, T), then there is a partition of unity subordinate to U . [Hint: Use (xiii) above and Exercises 10.3 #28(i).] (xv)* Every metrizable space is paracompact. (xvi) If (X, τ ) is paracompact Hausdorﬀ space and (Y, τ 1) is a compact Hausdorﬀ space, then the product space (X, τ ) × (Y, τ 1) is paracompact Hausdorﬀ;. (xvii) If f : (X, τ ) → (Y, τ 1) is a continuous closed surjective map of the paracompact Hausdorﬀ space (X, τ ) onto the Hausdorﬀ space (Y, τ 1), then (Y, τ 1) is paracompact. [Hint: Use (xii) above.] (xviii) Every countably compact paracompact space is compact. (xix) A topological space (X, τ ) is said to be metacompact if for any open cover {Ai : i ∈ I} of (X, τ ), there is a reﬁnement {Bj : j ∈ J} which is also an open cover of (X, τ ) with the property that every point x ∈ X is contained in only ﬁnitely many sets Bj, j ∈ J . Prove that every paracompact space is metacompact and the product of a metacompact Hausdorﬀ space and a compact Hausdorﬀ space is metacompact. 10.3. TYCHONOFF’S THEOREM 311 30. Using previous exercises, prove the correctness of the picture below: 312 CHAPTER 10. TYCHONOFF’S THEOREM Cellularity 31. Let (X, τ ) be a topological space, the cellularity of the topological space (X, τ ) is the cardinal number 8 c(X) given by c(X) = ℵ0 + sup{card U : U is a set of pairwise disjoint open sets in (X, τ ).} The density of (X, τ ) is the cardinal number d(X) given by d(X) = ℵ0 + min{card S :
S ⊆ X and S is dense in X}. The spread of (X, τ ) is the cardinal number s(X) given by s(X) = ℵ0 + sup{card (D) : D is a discrete subspace of X}. The Lindelöf degree, L(X), of (X, τ ) is the smallest inﬁnite cardinal number ℵ such that every open cover of X has a subcover of cardinality ℵ. Of course L(X) = ℵ0 if and only if (X, τ ) is a Lindelöf space. Write down and verify the obvious relations between cellularity, density, spread, Lindelöf degree, and weight of any topological space (X, τ ). G-bases 32. Recall9 the deﬁnition of poset in Deﬁnition 10.2.1. A subset A of the poset P is said to be coﬁnal in P if for each x ∈ P , there exists an a ∈ A such that x a. A subset A of the poset P is said to be coinitial in P if for each x ∈ P , there exists a b ∈ A such that b x. A subset A of P is said to be bounded if there exists an x ∈ P such that for all a ∈ A, a x. A poset P is said to be lower complete if for every subset A of X has a greatest lower bound. The coﬁnality of a poset P , denoted by cf(P ) is the least of the cardinalities of the coﬁnal subsets of P . 8There are numerous cardinal invariants (cardinal functions) associated with each topological space. We have introduced only 4 of these in this exercise. For more detailed discussion, see Juhász [222] and Chapter 1 of Kunen and Vaughan [248]. 9This exercise is inspired by the beautiful manuscript, Banakh [28]. Most of the material here appears in that manuscript of Taras Banakh (1968– ). 10.3. TYCHONOFF’S THEOREM 313 Let P and Q be posets. The poset Q is said to be Tukey dominated by P , denoted by Q T P , if there exists a function f : P → Q that maps every conﬁnal subset of P to a coﬁnal subset of Q. The posets P and Q are said to be Tukey equivalent if P T Q and Q T P . If (X, τ ) is a topological space, for each point x ∈ X the set Nx of all neighbourhoods in (X, τ ) of x is a poset if we give it the partial order of reverse inclusion; that is, if N1, N2 ∈ Nx then N2 N1 precisely when N1 ⊆ N2. So Nx is a directed lower complete poset. (A poset P is said to be directed if for every a, b ∈ P , there exists a c ∈ P such that a c and b c.) For x ∈ X, the poset Nx is Tukey dominated by ω if and only if the topological space (X, τ ) is ﬁrst countable at the point x. Let P be a poset. The topological space (X, τ ) is said to have a neighbourhood P -base at the point x ∈ X if the poset Nx of all neighbourhoods of x is Tukey dominated by P . Since Nx is lower complete, this happens if and only if at each x ∈ X, the space (X, τ ) has a neighbourhood base {Uα[x] : α ∈ P } If a topological space (X, τ ) such that Uβ[x] ⊆ Uα[x] for all α β ∈ P . has a neighbourhood P -base {Uα : α ∈ P } at each x ∈ X, then these neighbourhood bases can be encoded by the set {Uα : α ∈ P } of the entourages Uα = {(x, y) ∈ X × X : y ∈ Uα[x]}, α ∈ P . Such a family {Uα : α ∈ P } is said to be a P -based topological space. A topological space (X, τ ) has a neighbourhood P -base at each point if and only if it has a P -base. Banakh [28], Banakh and Leiderman [29], Gabriyelyan and K¸akol [144], Leiderman et al. [257], and Gabriyelyan et al. [145] study topological spaces, topological groups and topological vector spaces with an ωω-base. (Note that by ωω we mean the uncountable cardinal number of the set NN with co-ordinatewise partial ordering , given by f g when f (n) g(n), for f, g : N → N and n ∈ N, not the countable ordinal number.) In some literature these are known as topological spaces with a G-base. (i) Let P and Q be posets and f : P → Q. Verify that Q is Tukey dominated by P if and only if f −1(B) is bounded in P for every bounded subset B of P . (ii) Verify that a poset is Tukey dominated by ω if and only if P is a directed poset of countable coﬁnality. 314 CHAPTER 10. TYCHONOFF’S THEOREM (iii) Let P and Q be posets. If Q has a greatest element, verify that P is Tukey dominated by Q. (iv) Let P and Q be posets. If both P and Q have greatest elements, the P and Q are Tukey equivalent. (v) Verify that a topological space (X, τ ) is ﬁrst countable if and only if it has an ω-base. (vi) Verify that every topological space with an ω-base has an ωω-base. Deduce that every metrizable space has an ωω- base. Banach spaces, Dual Spaces, Weak Topologies, and Reﬂexivity 33. Let N be a normed vector space over K, which is the ﬁeld R of real numbers or C of complex numbers, with norm \|\| \|\|. A linear map from the underlying vector space of N into K is said to be a linear functional. Denote the set of all continuous linear functionals from N into K by N ∗. If φ, φ1, φ2 ∈ N ∗ and λ ∈ K, deﬁne (φ1 + φ2)(x) = φ1(x) + φ2(x) and (λφ)(x) = λ(φ(x)). With these operations N ∗ is a vector space over K. Deﬁne a norm \|\| \|\|op on N ∗ by \|\|φ\|\|op = sup{\|φ(x)\| : x ∈ N , \|\|x\|\| 1}, for φ ∈ N ∗. (op is short for operator.) (i) Verify that N ∗, with \|\| \|\|op, is indeed a normed vector space. (The normed vector space N ∗ is called the dual space of N .) (ii) Verify that with this norm, N ∗ is in fact a Banach space. [Hint: Let φn be a Cauchy sequence of members of N ∗. For each x ∈ N , consider the sequence φn(x) in the ﬁeld K of scalars. Use the fact that K is a complete metric space.] (iii)** Verify each of the following: (a) if N is Rn, n ∈ N, then N ∗ is isometrically isomorohic to Rn; (b) if N is p for 1 < p < ∞, then N ∗ is isometrically isomorphic to q, where p + 1 1 q = 1; (c) if N is 2, then N ∗ is isometrically isomorphic to 2; (d) if N is c0, then N ∗ is isometrically isomorphic to 1; (e) if N is 1, then N ∗ is isometrically isomorphic to ∞. 10.3. TYCHONOFF’S THEOREM 315 (iv) Consider the double dual N ∗∗ of N ; that is, N ∗∗ = (N ∗)∗. Verify that the natural map Γ : N → N ∗∗ given by Γ(x)(φ) = φ(x), for x ∈ N , φ ∈ N∗ is one-to-one, linear and norm preserving; that is, \|\|Γ(x)\|\|op = \|\|x\|\|, for all x ∈ N . Deduce that if N is a Banach space, then the Banach spaces N and Γ(N ) are isomorphic as Banach spaces; that is, the map Γ : N → Γ(N ) ⊆ N ∗∗ is one-to-one and onto and an isometric isomrphism of N onto Γ(N ). (v) A normed vector space is said to be reﬂexive if in the notation above, Γ(N ) = N ∗∗. Deduce that if N is reﬂexive, then it is a Banach space and N and N ∗∗ are isometric Banach spaces. (vi) Let N be a normed vector space over the scalar ﬁeld K which is R or C, as usual. Then the norm on N induces a topology on N which is called the strong topology. Let N ∗ be the dual space of N , which we have seen is a Banach space. Obviously the strong topology is such that, by deﬁnition, every φ : N → K, where φ ∈ N ∗, is continuous. However, there may be coarser topologies on the vector space N such that all such φ are continuous. The coarsest topogy τ on the vector space N such that for each φ ∈ N ∗, the linear functional (N, τ ) → K is continuous is called the weak topology. The coarsest topology τ ∗ on the vector space N ∗ such that every linear functional Γ(x) : N ∗ → K, given by Γ(x)(φ) = φ(x), for each x ∈ N and φ ∈ N ∗, is called the weak-topology. Verify from the deﬁnitions that the weak-topology on N ∗ is coarser than the weak topology on N ∗ but the two topologies coincide if N is a reﬂexive Banach space. 316 CHAPTER 10. TYCHONOFF’S THEOREM (vii) [Banach-Alaoglu Theorem] (Leonidas Alaoglu (1914–1981) was a Canadian mathematician who extended Banach’s result from separable spaces to the general case.) If N is a normed vector space, then the closed unit ball of N ∗ is weak-compact; that is, compact in the weak-topology on N ∗. (This result should be contrasted with that in Exercises 8.3 #3 (vi).) Prove this theorem by verifying each of the following steps. (a) Let B∗ = {φ : φ ∈ N ∗, \|\|φ\|\|∗ 1} be the closed unit ball of N ∗, where \|\| \|\|∗ is the norm on N ∗. Then each linear functional φ ∈ B∗ maps the closed unit ball B = {x : x ∈ N, \|\|x\|\| 1} of N into the disc D = {c : c ∈ K, \|c\| 1}. (b) So we can identify B∗ with a subset of the product space DB. the space (c) We can readily check that the weak-topology on B∗ is in fact the subspace topology from the product topology on DB. (d) Indeed, B∗ is a closed subspace of DB. (e) As D is compact, Tychonoﬀ ’s Theorem 10.3.4 implies that DB is compact. (e) So the closed subspace B∗ of DB is compact; that is, B∗ is weak- compact. (viii) Let K be the real number ﬁeld R or the complex number ﬁeld C and V, W vector spaces over K. A linear operator from V to W is a map φ : V → W such that φ(λx + µy) = λφ(x) + µφ(y), for all x, y ∈ V and all λ, µ ∈ K. If V and W are normed vector spaces, a linear operator φ : V → W is said to be bounded if there exists a positive real number M such that \|\|φ(x)\|\| M \|\|x\|\| for every x ∈ V . Prove that a linear operator φ : V → W is bounded if and only if it is continuous. The smallest such M is called the operator norm and denoted \|\| \|\|op, and the set B(V, W ) of all linear operators from V to W with this norm \|\| \|\|op is a normed vector space. (Observe that the norm \|\| \|\|op depends on the norms on V and W .) The normed vector space (B(V, W ), \|\| \|\|op) is called the space of bounded linear operators. Deﬁne bounded linear functional and deduce from what we just proved that a linear functional φ : V → K is bounded if and only if it is continuous. 10.3. TYCHONOFF’S THEOREM 317 (ix) Let V = C[0, 1] the Banach space of all continuous functions f : [0, 1] → R 1 1 2 0 \|f (x)\|2dx with the norm given by \|\|f \|\| = . Deﬁne a linear functional φ : V → R by φ(f ) = f (0), for all f ∈ C[0, 1]. Verify (a) V is a normed vector space; (b) V is a Banach space; (c) φ is a linear functional; (d) by considering functions fn : [0, 1] → R such that fn(0) = n but 1 0 (fn(x))2 = 1, for each positive integer n, that φ is not a bounded linear functional. (x)* (One-dimensional version of the Hahn-Banach Theorem over R The Hahn-Banach Theorem was proved independently in the 1920s by the Austrian mathematician Hans Hahn (1879–1934) and the Polish mathematician Stefan Banach (1892–1945).) Prove the following lemma. Lemma. Let B be a normed vector space, E
a vector subspace of B, and f : E → R a bounded linear functional. Then for any x ∈ B \ E, there exists a linear functional f1 : E1 = span{E, x1} → R that extends f (that is, f (x) = f1(x), for all x ∈ E) and satisﬁes \|\|f \|\|op = \|\|f1\|\|op. (If X is a subset of a vector space L, span(L) denotes the smallest vector subspace of L which contains X.) [Hint. If \|\|f \|\|op = 0, the result is seen trivially to be true. So we can, without loss of generality, assume \|\|f \|\|op = 1. Now if x ∈ E1, then x = λx1 + y, for λ ∈ R and y ∈ E. To deﬁne f1 as an extension of f it suﬃces to choose an appropriate value of f1(x1) – by appropriate we mean a value such that \|\|f \|\|op = \|\|f1\|\|op. Let us put f1(x1) = a1, so that f1(λx1 + y) = λa1 + f (y). It therefore suﬃces to choose a1 such that \|f1(x)\| \|\|x\|\|, for all x ∈ E1; that is, −\|\|λx1 + y\|\| λa1 + f (y) \|\|λx1 + y\|\| , for all λ ∈ R and y ∈ E. − x1 + This holds for λ = 0 by hypothesis on f , so we can assume λ = 0. This allows us to rewrite the above inequality as y λ Equivalently − \|\|x1 + z\|\| − f (z) a1 \|\|x1 + z\|\| − f (z), for all z ∈ E. Show that for any z1, z2 ∈ E , for all λ ∈ R and y ∈ E. x1 + − f − f a1 y λ y λ y λ −\|\|x1 + z1\|\| − f (z1) \|\|x1 + z2\|\| − f (z2). 318 CHAPTER 10. TYCHONOFF’S THEOREM Observe that this implies −∞ < sup z1∈E [−\|\|x1 + z1\|\| − f (z1)] inf z2∈E [\|\|x1 + z1\|\| − f (z2)] < ∞. So we can choose for a1 any value between sup z1∈E [−\|\|x1 + z1\|\| − f (z1)] and inf z2∈E [\|\|x1 + z1\|\| − f (z2)] .] (xi)* [Hahn-Banach Theorem] Let B be a normed vector space over a ﬁeld K, where K is R or C, and E a vector subspace of B. If f : M → K is a bounded linear functional, then there exists a bounded linear function f1 : B → K that extends f . Using (x) above and Zorn’s Lemma, prove the Hahn-Banach Theorem. [Hint. For convenience we discuss only the case that K = R. We saw in (x) above how to extend a bounded linear functional from E to a vector space, one dimension greater. We could extend once more to a vector space of dimension 2 greater. Indeed we could do such an extension a ﬁnite number of times. But to extend to a space of uncountable dimension greater, it is necessary to use the Axiom of Choice or its equivalent, Zorn’s Lemma. Let B consist of all pairs (N, fN ) such that (a) N is a vector subspace of B that contains E; (b) fN is a bounded linear functional on N ; (c) fN extends f ; that is fN (x) = f (x) for all x ∈ E; (d) \|\|fN \|\|op = \|\|f \|\|op. Now put a partial order on B as follows: (N, fN ) (N , fN ) if N ⊆ N and fN extends fN on N . Complete the proof by considering a totally ordered subset of B, observing that it has an upper bound, deducing by Zorn’s Lemma that the partial ordering has a maximal element (NM , fM ) and using (x) to show (NM , fM ) cannot be a maximal element unless BM = B.] 10.3. TYCHONOFF’S THEOREM 319 (xii) Let B be a normed vector space over R and 0 = x1 ∈ B. Using the Hahn-Banach Theorem show that there exists a bounded linear functional f1 : B → R with \|\|f \|\|op = 1 and f1(x1) = \|\|x1\|\|. [Hint. Let E be the vector subspace of B spanned by x1. Deﬁne a linear functional f on E by f (λx1) = λ\|\|x1\|\|. Show that f is a bounded linear functional and \|\|f \|\|op = 1. Then extend f to all of B using the Hahn-Banach Theorem.] (xiii) Let B be a normed vector space and E a proper closed subspace of B. Then there exists a bounded linear functional f : B → R with \|\|f \|\|op = 1 and f (x) = 0, for all x ∈ E. (This result generalizes that in (xii) above.) (xiv) Prove the following statement: Let E be a normed vector space. If E∗ is separable, then E is separable. [Hint. Noting that the unit sphere in E∗ is separable, let {fn ∈ E∗ : n ∈ N} be a countable dense subset of this unit sphere. Verify that for each n ∈ N, there exists xn ∈ E with \|\|xn\|\| = 1 such that \|fn(xn)\| 1 2 . Then let W be the closed linear span of {xn : n ∈ N}. The aim is to show that W = B. Suppose to the contrary that W is a proper closed subspace of B. By (xiii) there exists an f ∈ B∗ with \|\|f \|\|op = 1 and f (x) = 0, for all x ∈ W . So f (xn) = 0, for all n ∈ N. Show that this implies that for all n ∈ N 1 2 \|fn(xn)\| = \|fn(xn) − f (xn)\| \|\|fn − f \|\|op\|\|xn\|\| = \|\|fn − f \|\|op. But this contradicts the assumption that {fn : n ∈ N} is dense in the unit sphere of B∗.] 320 CHAPTER 10. TYCHONOFF’S THEOREM (xv) Let (N, \|\| \|\|N ) be a normed vector space and E a closed vector subspace of N . We deﬁne an equivalence relation ∼ on N by x ∼ y if x − y ∈ E, x, y ∈ N and deﬁne N/E to be the set of all these equivalence classes. We deﬁne the map g : N → N/E by g(x) is the equivalence class of x. Deﬁne \|\|z\|\|N/E = inf x∈N {\|\|x\|\|N : g(x) = z}. Prove the following: (a) Deﬁne the natural vector space structure on N/E induced by the vector space structure on N . Verify that N/E is indeed a vector space; (b) \|\| \|\|N/E is a norm on the vector space N/E; (c) g is an open continuous linear operator; (d) if N is a Banach space, then (N/E, \|\| \|\|N/E) is a Banach space. The normed vector space \|\|(N/E, \|\| \|\|N/E) is said to be the quotient space of N by E. (See also quotient (topological) space in Deﬁnitions 11.1.1 and quotient (topological) group in Proposition A5.2.16.) (xvi) Let N be a normed vector space, E a subset of N . Then E⊥ is deﬁned to be {f ∈ N ∗ : f (x) = 0, for all x ∈ E}, and is called the annihilator of E. Prove that E⊥ is a closed vector subspace of N ∗ and that E⊥ = (span E)⊥. If F is a subset of N ∗, then F⊥ is deﬁned to be {x ∈ N : f (x) = 0, ∀f ∈ F } and is called the pre-annihilator of F . Prove also that F⊥ is a closed vector subspace of N and F⊥ = (span F )⊥. 10.3. TYCHONOFF’S THEOREM 321 (xvii)* Prove the following statement: Let N be a normed vector space and E a vector subspace of N . Then (a) E∗ is isometrically isomorphic to N ∗/E⊥; (b) If E is closed, then (N/E)∗ is isometrically isomorphic to E⊥. (Note that this result for Banach spaces is the analogue of Corollary A5.9.6 and Corollary A5.9.7 in Pontryagin Duality of topological groups.) [Hint. (a) Let Θ : N ∗ → N ∗/E⊥ be the natural mapping. Consider f ∈ E∗. By the Hahn-Banach Theorem there exists a bounded linear functional f1 : N → R such that f1 is an extension of F and \|\|f1\|\|op = \|\|f \|\|op. Now prove that the map φ : E∗ → N ∗/E⊥ given by φ(f ) = Θ(f1) is an isometry of E∗ onto N ∗/E⊥. (b) Let Γ : N → N/E be the natural quotient mapping. Deﬁne the map δ : (N/E)∗ → E∗ by δ(f ) = f ◦ Γ. Verify that δ is linear and δ(f )(x) = 0, for all x ∈ E. Deduce from this that δ maps (N/E)∗ into E⊥. Now prove that δ is an isometry of (N/E)∗ onto E⊥.] (xviii) Prove the following statement: Let B be any Banach space and let Φ : B → B∗∗ be the canonical linear mapping given by Φ(x)(γ) = γ(x), for all γ ∈ B∗ and x ∈ B. Then Φ maps B isometrically onto the subspace Φ(B) of B∗∗. (xix) Prove the following result: A Banach space B is reﬂexive if and only if B∗ is reﬂexive. [Hint. Firstly consider the case that B is reﬂexive. As B is reﬂexive, linear operator Φ : B → B∗∗ is an isometry. Using (xviii) the canonical above, to prove that B∗ is reﬂexive, it suﬃces to show that the canonical bounded linear operator Ψ from B∗ to B∗∗∗ is surjective. Let γ ∈ B∗∗∗. So γ ∈ (Φ(B))∗; that is, γ is a bounded linear functional from Φ(B) to R. Let fγ : B → R be deﬁned by fγ(b) = γ(Φ(b)), for all b ∈ B. Verify that fγ ∈ B∗ and that Ψ(fγ) = γ. This shows that Ψ has the required properties. Conversely, suppose that B is not reﬂexive. Then Φ(B) is a proper subspace of B∗∗. Use (xiii) above to obtain a contradiction.] 322 CHAPTER 10. TYCHONOFF’S THEOREM (xx) Deduce the following theorem using the Banach-Alaoglu Theorem: The Banach space B is reﬂexive if and only if its closed unit ball T is weakly compact. [Hint. First use the Banach-Alaoglu Theorem to prove that the closed unit ball in a reﬂexive Banach space B is weakly compact. Observe that if B is a reﬂexive Banach space, then B = B∗∗ and the weak topology on B coincides with the weak-topology on (B∗)∗ = B. Deduce that T is weakly compact. Conversely assume that T is weakly compact and let T be the closed unit ball in B∗∗. The embedding of B in B∗∗ is weak-weak continuous, so the image of T under this embedding is weak-compact in B∗∗. Using the fact that the image of T is weak-dense in T , we can see the image of T is in fact T . From this it follows that the image of B in B∗∗ equals B∗∗, that is B is reﬂexive.] (xxi) A Banach space B is said to be weakly compactly generated or WCG if it has a subset S which is weakly compact (that is, compact in the weak topology on B) such that the closed linear span of S is B. Prove using Exercises 8.3 #3(vi) that every separable Banch space is a WCG space and using (xx) above that every reﬂexive Banach space is also a WCG space. (xxii) Let B be a Banach space and E a closed subspace of B. By considering the quotient Banach space B/E, show that for any x1 ∈ B \ E there exists a bounded linear functional fx1 : B → R such that fx1(y) = 0, for all y ∈ E but fx1(x1) = 0. Deduce that if E is a closed vector subspace of the Banach space B, then E is a closed subspace of B with its weak topology. (xxiii)* Using (xx) prove that every closed subspace of a reﬂexive Banach space is a reﬂexive Banach space. [Hint. Let B be a reﬂexive Banach space and E a closed subspace of B. Let T and S be the closed unit balls in B and E, respectively. As B is reﬂexive, T is weakly compact. Using this and (xxii) above, show that S is weakly compact.] 10.3. TYCHONOFF’S THEOREM 323 (xxiv)** Prove that a Banach space B is separable if and only if the closed unit ball of B∗ is weak*-metrizable. (This result complements that of Exercises 8.3 #3 (vi).) (xxv) In the 1930s the Polish mathematicians Stefan Banach and Stanisław Mazur (1905–1981) stated a question, which has become known as the Separable Quotient Problem and is still unanswered. Does every inﬁnite- dimensional Banach space have a quotient space which is an inﬁnite- dimensional separable Banach space? This has been proved to be true for a variety of special cases, such as w
hen the inﬁnite-dimensional Banach space B is reﬂexive. (For other special cases, see Argyros et al. [14].) To verify the reﬂexivity special case, check that the following statements are true: (a) Every inﬁnite-dimensional Banach space has a subspace which is an inﬁnite-dimensional separable Banach space. (b) Every closed subspace of a reﬂexive Banach space is reﬂexive. (c) If a Banach space B is such that B∗ is separable, then B is separable. (d) If a separable reﬂexive Banach space B, then B∗ is separable. (e) If B is any inﬁnite-dimensional reﬂexive Banach space, then its dual space B∗ has an inﬁnite-dimensional separable reﬂexive Banach subspace E which implies that B∗∗ has an inﬁnite-dimensional separable (reﬂexive) quotient Banach space E∗. (f ) If B is an inﬁnite-dimensional reﬂexive Banach space, then B has an inﬁnite-dimensional separable (reﬂexive) quotient Banach space. 324 CHAPTER 10. TYCHONOFF’S THEOREM (xxvi) A Banach space B is said to an Asplund space, named after Edgar Asplund (1931–1974), if every separable closed subspace of B has a separable dual. (a) Using (xxv) above, verify that every reﬂexive Banach space is an Asplund space. (b) Verify that every closed subspace on an Asplund space is an Asplund space. (c) Noting that the dual space of the separable Banach space 1 is the nonseparable space ∞, we see that not every separable Banach space is an Asplund space. Deduce that not every weakly compactly generated space is an Asplund space. (d) Prove that a separable Banach space B is an Asplund space if and only if its dual B∗ is a separable Banach space. (e) Verify that every Banach space which is a quotient space of an Asplund space is an Asplund space. (f ) Verify that every ﬁnite product of Asplund spaces is an Asplund space. Remarks. (α) As a generalization of (f ) we mention that being an Asplund space is a three space property; that is, if B is a Banach space which has a closed vector subspace E such that E and the quotient Banch space B/E are Asplund spaces, then B is an Asplund space. (See Exercises A5.12 #5 where the three space property was introduced.) This result is Theorem 4.11a of Castillo and González [73]. If B is a Banach space and the dual space B∗ is a WCG space, then (β) B∗ is an Asplund space. (See Theorem 4.11b of Castillo and González [73].) (g) Verify that if B is an inﬁnite-dimensional Asplund space, then its dual space B∗ has a quotient space which is an inﬁnite-dimensional separable Banach space. (h) Verify that every reﬂexive Banach space (i) is an Asplund space and (ii) is the dual of an asplund space. Deduce that (g) above includes as a special case (xxv)(f ). 10.3. TYCHONOFF’S THEOREM 325 (i) Prove that if the Banach space B has a subspace E, where E is an inﬁnite-dimensional Asplund space, then the dual space B∗ has a quotient space which is an inﬁnite-dimensional separable Banach In particular, this is the case if E is a reﬂexive Banach space. space. (j) Verify from the above that if B is a Banach space which contains as a subspace c0 or p, for 1 < p < ∞, or any Hilbert space, then B∗ has an inﬁnite-dimensional separable quotient Banach space. (xxvii) Using Exercises 8.3 #3(xi) prove that if the Banach space B has Banach subspaces B1 ⊂ B2 ⊂ · · · ⊂ Bi ⊂ · · · ⊂ . . . with each Bi+1/Bi having dimension 1, and ∞ an inﬁnite-dimensional separable quotient Banach space. i=1 Bi dense in B, then B has (xxviii)(a) Let B be an inﬁnite-dimensional Banach space such that its dual space B∗ has a separable inﬁnite-dimensional Banach quotient space. Using (xviii) show that B∗∗∗ has a separable inﬁnite-dimensional Banach quotient space. Deduce that B∗∗∗∗∗, B∗∗∗∗∗∗∗, etc. have inﬁnite-dimensional separable quotient Banach spaces. (b) Let B be a Banach space with an inﬁnite-dimensional Banach subspace E. If E∗ has a separable inﬁnite-dimensional Banach quotient space, show that B∗∗∗ has a separable inﬁnite-dimensional Banach quotient space. Deduce that B∗∗∗∗∗, B∗∗∗∗∗∗∗, etc. have inﬁnite-dimensional separable quotient Banach spaces. (c) Deduce from the above that if a Banach space B has a subspace E which is an inﬁnite-dimensional Asplund space, then the dual spaces B∗∗∗, B∗∗∗∗∗, etc have inﬁnite-dimensional separable quotient Banach spaces. 326 CHAPTER 10. TYCHONOFF’S THEOREM 10.4 Stone-˘Cech Compactiﬁcation Let (X, τ ) be a topological space, (βX, τ ) a 10.4.1 Deﬁnition. compact Hausdorﬀ space and β : (X, τ ) −→ (βX, τ ) a continuous mapping, then (βX, τ ) together with the mapping β is said to be the Stone-˘Cech compactiﬁcation of (X, τ ) if for any compact Hausdorﬀ space (Y, τ ) and any continuous mapping φ : (X, τ ) −→ (Y, τ ), there exists a unique continuous mapping Φ : (βX, τ ) −→ (Y, τ ) such that Φ ◦ β = φ; that is, the diagram below commutes: (X, τ ) β ..................................................................................................................................................................................... ............ (βX, τ ) ............................................................................................................................................................................................................................................................................................................................................................................ ............ φ ................................................................................................................................................................... ............ Φ (Y, τ ). WARNING The mapping β is usually not surjective, so β(X) is usually not equal to βX. 10.4.2 Remark. Those familiar with category theory should immediately recognize that the existence of the Stone-˘Cech compactiﬁcation follows from the Freyd Adjoint Functor Theorem using the forgetful functor from the category of compact Hausdorﬀ spaces and continuous functions to the category of topological spaces and continuous functions. For a discussion of this see MacLane [269], Freyd [140]10 While the Stone-˘Cech compactiﬁcation exists for all topological spaces, it assumes more signiﬁcance in the case of Tychonoﬀ spaces. For the mapping β is an embedding if and only if the space (X, τ ) is Tychonoﬀ. The “only if ” part of this is clear, since the compact Hausdorﬀ space (βX, τ ) is a Tychonoﬀ space and so, therefore, is any subspace of it. 10Peter Freyd’s book “Abelian categories: An introduction to the theory of functors” is available as a free download at various sites including http://www.emis.de/journals/TAC/reprints/articles/3/tr3.pdf. 10.4. STONE- ˘CECH COMPACTIFICATION 327 We now proceed to prove the existence of the Stone-˘Cech compactiﬁcation for Tychonoﬀ spaces and of showing that the map β is an embedding in this case. Let (X, τ ) and (Y, τ ) be Tychonoﬀ spaces and F(X) and 10.4.3 Lemma. F(Y ) the family of all continuous mappings of X and Y into [0, 1], respectively. Further let eX and eY be the evaluation maps of X into f ∈F(X) If and Y into ∼= Ig ∼= [0, 1], for each f and g. If φ is any continuous mapping of X into Y , then there exists a continuous mapping Φ of g∈F(Y ) Ig such that Φ ◦ eX = eY ◦ φ; that is, the diagram below commutes. g∈F(Y ) Ig, respectively, where If f ∈F(X) If into (X, τ ) .............................................................................................................................. ............φ (Y, τ ) eX ............................................................................................................................................................. ............ eY ............................................................................................................................................................. ............ Ig f ∈F(X) If Further, Φ(eX (X)) ⊆ eY (Y ). ...................................................... ............ Φ g∈F(Y ) = Let f ∈F(X) xf ∈ g∈F(Y ) yg, f ∈F(X) If . Deﬁne Φ Proof. where yg is deﬁned as follows: as g ∈ F(Y ), g is continuous map from (Y, τ ) into [0, 1]. So g ◦ φ is a continuous map from (X, τ ) into [0, 1]. Thus g ◦ φ = f , for some f ∈ F(X). Then put yg = xf , for this f , and the mapping Φ is now deﬁned. f ∈F(X) xf To prove continuity of Φ, let U = f ∈F(X) xf ) = g∈F(Z) Ug be a basic open set containing Φ( g∈F(Y ) yg. Then Ug = Ig for all g ∈ F(Y ) \ {gi1, . . . , gin}, for gi1, . . . , gin. Put fi1 = gi1 ◦ φ, fi2 = gi2◦, . . . , fin = gin ◦ φ. Now deﬁne V = f ∈F(X) Vf , where Vf = If , for some f ∈ F(X) \ {fi1, fi2, . . . , fin}, and . Clearly Vfi1 f ∈F(X) xf ∈ V and Φ(V ) ⊆ U . So Φ is continuous. , . . . , Vfin = Ugi1 = Ugi2 = Ugin , Vf12 To see that the diagram commutes, observe that Φ(eX (x)) = Φ f (x) = g(φ(x)), for all x ∈ X. f ∈F(X) g∈F(Y ) So Φ ◦ eX = eY ◦ φ. Finally as Φ is continuous, Φ(eX (X)) ⊆ eY (Y ), as required. 328 CHAPTER 10. TYCHONOFF’S THEOREM Let Φ1 and Φ2 be continuous mappings of a topological 10.4.4 Lemma. space (X, τ ) into the Hausdorﬀ space (Y, τ ). If Z is a dense subset of (X, τ ) and Φ1(z) = Φ2(z) for all z ∈ Z, then Φ1 = Φ2 on X. Suppose Φ1(x) = Φ2(x), for some x ∈ X. Then as (Y, τ ) is Proof. Hausdorﬀ, there exist open sets U Φ1(x) and V Φ2(x), with U ∩ V = Ø. So Φ−1 1 (U ) ∩ Φ−1 As Z is dense in (X, τ ), there exists a z ∈ Z such that z ∈ Φ−1 2 (V ). So Φ1(z) ∈ U and Φ2(z) ∈ V . But Φ1(z) = Φ2(z). So U ∩ V = Ø, which is a contradiction. 2 (V ) is an open set containing x. 1 (U ) ∩ Φ−1 Hence Φ1(x) = Φ2(x), for all x ∈ X. 10.4. STONE- ˘CECH COMPACTIFICATION 329 Let (X, τ ) be any Tychonoﬀ space, F(X) the family 10.4.5 Proposition. of continuous mappings of (X, τ ) into [0, 1], and eX the evaluation map of (X, τ ) into ∼= [0, 1]. Put (βX, T ) equal to eX (X) with the subspace topology and β : (X, τ ) −→ (βX, τ ) equal to the mapping eX . Then (βX, τ ) together with the mapping β is the Stone-˘Cech compactiﬁcation of (X, τ ). f ∈F(X) If , where each If Proof. Firstly observe that (βX, τ ) is indeed a compact Hausdorﬀ space, as it i
s a closed subspace of a compact Hausdorﬀ space. Let φ be any continuous mapping of (X, τ ) into any compact Hausdorﬀ space (Y, τ ). We are required to ﬁnd a mapping Φ as in Deﬁnition 10.4.1 so that the diagram there commutes and show that φ is unique. Let F(Y ) be the family of all continuous mappings of (Y, τ ) into [0, 1] and eY the evaluation mapping of (Y, τ ) into g∈F(Y ) Ig, where each Ig ∼= [0, 1]. By Lemma 10.4.3, there exists a continuous mapping Γ : f ∈F(X) If −→ g∈F(Y ) Ig, such that eY ◦ φ = Γ ◦ eX , and Γ(eX (X)) ⊆ eY (Y ); that is, Γ(βX) ⊆ eY (Y ). As (Y, τ ) is a compact Hausdorﬀ space and eY is one-to-one, we see that eY (Y ) = eY (Y ) and eY : (Y, τ ) −→ (eY (Y ), τ ) is a homeomorphism, where : (eY (Y ), τ ) −→ (Y, τ ) is a τ homeomorphism. Put Φ = e−1 Y ◦ Γ so that Φ is a continuous mapping of (βX, τ ) into (Y, τ ). is the subspace topology on eY (Y ). So e−1 Y Further, for any x ∈ X Φ(β(x)) = Φ(eX (x), = e−1 = e−1 = φ(x). Y (Γ(eX (x))) Y (eY (φ(x))), as eY ◦ φ = Γ ◦ eX Thus Φ ◦ β = φ, as required. Now suppose there exist two continuous mappings Φ1 and Φ2 of (βX, τ ) into (Y, τ ) with Φ1 ◦ β = φ and Φ2 ◦ β = φ . Then Φ1 = Φ2 on the dense subset β(X) of (βX, τ ). So by Lemma 10.4.4, Φ1 = Φ2. Hence the mapping Φ is unique. 330 CHAPTER 10. TYCHONOFF’S THEOREM 10.4.6 Remark. implying that for each (X, τ ) there is a unique (βX, τ ). The next Proposition indicates in precisely what sense this is true. However we ﬁrst need a lemma. In Deﬁnition 10.4.1 referred to the Stone-˘Cech compactiﬁcation Let (X, τ ) be a topological space and let (Z, τ 1) together 10.4.7 Lemma. with a mapping β : (X, T ) −→ (Z, τ 1) be a Stone-˘Cech compactiﬁcation of (X, τ ). Then β(X) is dense in (Z, τ 1). Suppose β(X) is not dense in (Z, τ 1). Then there exists an element Proof. z0 ∈ Z \ β(X). As (Z, τ 1) is a compact Hausdorﬀ space, by Remark 10.3.28, it is a Tychonoﬀ space. Observing that Z \β(X) is an open set containing z, we deduce that there exists a continuous mapping Φ1 : (Z, τ 1) −→ [0, 1] with Φ1(z0) = 1 and Φ1(β(X)) = {0}. Also there exists a continuous mapping Φ2 : (Z, τ 1) −→ [0, 1 2 and Φ2(β(X)) = {0}. So we have the following diagrams which commute 2] with Φ2(z0) = 1 (X, τ ) β ............................................................................................................................................................................................... ............ (Z, τ 1) (X, τ ) β .......................................................................................................................................................................................................................................................................... ............ (Z, τ 1) ............................................................................................................................................................................................................................................................................................................................................................................ ............ φ ................................................................................................................................................................... Φ1 ............ [0, 1] ........................................................................................................................................................................................................................................................................................................................................................................................................ ............ φ ................................................................................................................................................................... ............ Φ3 Φ2 ......................................................................... ......................................................................... ............ [0, 1 2] e ............ [0, 1] where φ(x) = 0, for all x ∈ X and Φ3 is deﬁned by Φ3 = e ◦ Φ2, where e is the natural embedding of [0, 1 2] into [0, 1]. We see that the uniqueness of the mapping Φ in Deﬁnition 10.4.1 implies that Φ1 = Φ3, which is clearly false as Φ1(z0) = 1 and 2 . So our supposition is false and hence β(X) is dense in (Z, τ 1). Φ3(z0) = 1 10.4. STONE- ˘CECH COMPACTIFICATION 331 Let (X, τ ) be a topological space and (Z1, τ 1) 10.4.8 Proposition. together with a mapping β1 : (X, τ ) −→ (Z1, τ 1) a Stone-˘Cech compactiﬁcation of (X, τ ). If (Z2, τ 2) together with a mapping β2 : (X, τ ) −→ (Z2, τ 2) is also a Stone-˘Cech compactiﬁcation of (X, τ ) then (Z1, τ 1) ∼= (Z2, τ 2). Indeed, there exists a homeomorphism Θ : (Z1, τ 1) → (Z2, τ 2) such that Θ ◦ β1 = β2. (X, τ ) β1 ........................................................................................................................................................................................ ............ (Z1, τ 1) ............................................................................................................................................................................................................................................................................................................................................................................ ............ β2 ................................................................................................................................................................... ............ Θ (Z2, τ 2). As (Z1, τ 1) together with β1 is a Stone-˘Cech compactiﬁcation of (X, τ ) Proof. and β2 is a continuous mapping of (X, τ ) into the compact Hausdorﬀ space (Z2, τ 2), there exists a continuous mapping Θ : (Z1, τ 1) −→ (Z2, τ 2), such that Θ ◦ β1 = β2. Similarly there exists a continuous map Θ1 : (Z2, τ 2) −→ (Z1, τ 1) such that Θ1 ◦ β2 = β1. So for each x ∈ X, Θ1(Θ(β1(x))) = Θ1(β2(X)) = β1(x); that is, if on β1(X), which by idZ1 Lemma 10.4.7 is dense in (Z1, τ 1). So, by Lemma 10.4.4, Θ1 ◦ Θ = idZ1 is the identity mapping on (Z1, τ 1) then Θ1 ◦ Θ = idZ1 on Z1. Similarly Θ ◦ Θ1 = idZ2 on Z2. Hence Θ = Θ−1 1 means that Θ is a homeomorphism. and as both are continuous this Note that if if (X, τ ) is any Tychonoﬀ space and (βX, τ ) 10.4.9 Remark. together with β : (X, τ ) → (βX, τ ) is its Stone-˘Cech compactiﬁcation then the proof of Proposition 10.4.5 shows that β is an embedding. Indeed it is usual, in this case, to identify X with βX, and so regard (X, τ ) as a subspace of (βX, τ ). We, then, do not mention the embedding β and talk about (βX, τ ) as the Stone-˘Cech compactiﬁcation. 332 CHAPTER 10. TYCHONOFF’S THEOREM For the case that (X, τ ) is a compact Hausdorﬀ space, the 10.4.10 Remark. Stone-˘Cech compactiﬁcation of (X, τ ) is (X, τ ) itself. Obviously (X, τ ) together with the identity mapping into itself has the required property of a Stone-˘Cech it is the Stone-˘Cech compactiﬁcation. This compactiﬁcation. By uniqueness, could also be seen from the proof of Proposition 10.4.5 where we saw that for the compact Hausdorﬀ space (Y, τ ) the mapping eY : (Y, τ ) −→ (eY (Y ), τ ) is a homeomorphism. Stone-˘Cech compactiﬁcations of even nice spaces are usually 10.4.11 Remark. complicated. For example [0, 1] is not the Stone-˘Cech compactiﬁcation of (0, 1], since the continuous mapping φ : (0, 1] −→ [−1, 1] given by φ(x) = sin( 1 x) does not extend to a continuous map Φ : [0, 1] −→ [−1, 1]. Indeed it can be shown that the Stone-˘Cech compactiﬁcation of (0, 1] is not metrizable. We shall conclude this section be proving that the Stone-˘Cech compactiﬁcation of R, Q, (0, 1], and N each have cardinality 2c. Proposition 10.4.12 follows from our construction of the Stone-˘Cech compactiﬁcation in Proposition 10.4.5. Let (X, τ ) be a topological space, (K, τ 1) a 10.4.12 Proposition. compact Hausdorﬀ space, and θ : (X, τ ) → (K, τ 1) a continuous map. If for every continuous map φ : (X, τ ) → [0, 1] there exists a unique continuous map Φ : (K, τ i) → [0, 1] such that θ ◦ Φ = φ, then (K, τ 1) is the Stone-˘Cech compactiﬁcation of (X, τ ). Proof. Exercise. 10.4. STONE- ˘CECH COMPACTIFICATION 333 Let N have its discrete subspace topology in R and 10.4.13 Proposition. let βR be the Stone-˘Cech compactiﬁcation of R. If the closure in βR of N is denoted by clβRN , then clβRN is βN, the Stone-˘Cech compactiﬁcation of N. Let θ be any continuous map of N into [0, 1]. By the Tietze Extension Proof. Theorem 10.3.51, there is a continuous extension φ : R → [0, 1] of θ. Therefore there exists a continuous extension Φ : βR → [0, 1] of φ. So Φ\|clβR(N), the restriction of the map Φ to clβRN , maps clβRN to [0, 1] and is a continuous extension of θ. As N is dense in clβRN , this is the unique continuous extension. So by Proposition 10.4.12, clβRN is βN. 10.4.14 Remark. 10.4.13 that clβQN = βN. It can be proved in a similar manner to that in Proposition 10.4.15 Proposition. Then card βX card βN. Let (X, τ ) be any separable topological space. As (X, τ ) is separable, it has a a dense countable subspace (Y, τ 1). Proof. So there exists a continuous surjective map γ : N → (Y, τ 1). Thus there exists a continuous map Φ : βN → βX which has dense image in βX. But as the image is compact and βX is Hausdorﬀ, Φ(βN) = βX, from which the proposition immediately follows. 10.4.16 Proposition. card βN = card βQ = card βR. Proof. Remark 10.4.14. This follows immediately from Propositions 10.4.15 and 10.4.13 and 334 CHAPTER 10. TYCHONOFF’S THEOREM 10.4.17 Proposition. card βN = card βQ = card βR = 2c. Proof. By The Corollary 10.3.42 of the Hewitt-Marczewski-Pondiczery Theorem 10.3.41, the compact Hausdorﬀ space [0, 1]c is separable. So by Proposition 10.4.15, card [0, 1]c card βN; that is, 2c card βN. By the construction in Proposition 10.4.5, βN is a subspace of [0, 1]I , where I is the set of (continuous) functions from N to [0, 1
]. So card I = cℵ0 = c. Then card ([0, 1]I ) = cc = 2c. Thus card βN 2c. Hence card βN = 2c. The proposition then follows from Proposition 10.4.17. Let X be any unbounded subset of Rn, for any 10.4.18 Proposition. If τ is the euclidean subspace topology on X, then βX, the Stonen ∈ N. ˘Cech compactiﬁcation of (X, τ ), has a subspace homeomorphic to βN , and so card (βX) = 2c. Proof. Exercise. 10.4.19 Corollary. (X, τ ) is compact if and only if card (βX) = 2c. Let (X, τ ) be any closed subspace of Rn, n ∈ N. Then Proof. Exercise. 10.4.20 Corollary. cardinality m, then card (βX) = 22m . If (X, τ ) is an inﬁnite discrete topological space of Proof. Exercise. 10.4. STONE- ˘CECH COMPACTIFICATION 335 Exercises 10.4 1. Let (X, τ ) by a Tychonoﬀ space and (βX, τ ) its Stone-˘Cech compactiﬁcation. Prove that (X, τ ) is connected if and only if (βX, τ ) is connected. [Hint: Firstly verify that providing (X, τ ) has at least 2 points it is connected if and only if there does not exist a continuous map of (X, τ ) onto the discrete space {0, 1}.] 2. Let (X, τ ) be a Tychonoﬀ space and (βX, τ ) its Stone-˘Cech compactiﬁcation. If (A, τ 1) is a subspace of (βX, τ ) and A ⊇ X, prove that (βX, τ ) is also the Stone-˘Cech compactiﬁcation of (A, τ 1). [Hint: Verify that every continuous mapping of (X, τ ) into [0, 1] can be extended to a continuous mapping of (A, τ 1) into [0, 1]. Then use the construction of (βX, τ ).] 3. Let (X, τ ) be a dense subspace of a compact Hausdorﬀ space (Z, τ 1). If every continuous mapping of (X, τ ) into [0, 1] can be extended to a continuous mapping of (Z, τ 1) into [0, 1], prove that (Z, τ 1) is the Stone-˘Cech compactiﬁcation of (X, τ ). 4. Prove that card β(0, 1) = card β(0, 1] = card βP = 2c, where P is the topological space of irrational numbers with the euclidean topology. 5. Prove Proposition 10.4.18 and Corollary 10.4.19. 6. Is it true that if X is an unbounded set in an inﬁnite-dimensional normed vector space, then card (βX) 2c? 7. Using a similar method to that in the proof of Proposition 10.4.17, prove Corollary 10.4.20. 336 CHAPTER 10. TYCHONOFF’S THEOREM 10.5 Postscript At long last we deﬁned the product of an arbitrary number of topological spaces and proved the general Tychonoﬀ Theorem. (An alternative and more elegant proof of the Tychonoﬀ Theorem using the concept of a ﬁlter appears in Appendix 6.) We also extended the Embedding Lemma to the general case. This we used to characterize the Tychonoﬀ spaces as those which are homeomorphic to a subspace of a cube (that is, a product of copies of [0, 1]). Urysohn’s Lemma allowed us to obtain the following relations between the separation properties: T4 ⇒ T3 1 2 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0. Further, each of the properties compact Hausdorﬀ and metrizable imply T4. We have also seen a serious metrization theorem – namely Urysohn’s Metrization Theorem, which says that every regular second countable Hausdorﬀ space is metrizable. We introduced the Stone-˘Cech compactiﬁcation, which is a rich and serious topic of study in its own right. (See Gillman and Jerison [156]; Hindman and Strauss In order to show that the Stone-˘Cech compactiﬁcation is [187]; Walker [415].) huge, we proved other important resullts. One of these is the Hewitt-Marczewski- Pondiczery Theorem 10.3.41 which in particular showed the surprising fact that a product of c copies of R is separable. Extension Theorem 10.3.51 and stated it in a slightly more general form than is usual In a series of steps we proved the Tietze in most books. With the aid of these results we were able to prove, in Proposition 10.4.18, that if X is any unbounded subset of Rn, n ∈ N, then card (βX) = 2c. In particular, card βN = card βQ = card βR = card βP = 2c. In Exercises 10.3 #33 we have some beautful results from Banach space Theory. These include the Banach-Alaoglu Theorem, the Hahn-Banach Theorem, and a wealth of information on the duals of Banach spaces, the weak topology, the weak- topology on the dual space, reﬂexive Banach spaces, weakly compactly generated Banach spaces, the annihilator, quotients of Banach spaces, and the Open Question known as the Separable Quotient Problem. In particular, we show that every inﬁnite- 10.6. CREDIT FOR IMAGES 337 dimensional reﬂexive Banach space has a quotient Banach space which is a separable and inﬁnite-dimensional Banach space. It is known, but not proved here, that every inﬁnite-dimensional WCG space and every inﬁnite-dimensional dual of a Banach space has a quotient Banach space which is inﬁnite-dimensional and separable. 10.6 Credit for Images 1. Cliﬀord Hugh Dowker. Copyright information: http://www-history.mcs.st-andrews.ac.uk/Miscellaneous/Copyright.html 2. Edwin Hewitt. MFO. https://opc.mfo.de/detail?photo_id=1709 3. Edward Marczewski. MFO. https://opc.mfo.de/detail?photo_id=2747 4. Andrey Nikolayevich Tychonoﬀ (or Tikhonov). Photo taken by Konrad Jacobs. Copy from The Oberwolfach Photo Collection. The Oberwolfach Photo Collection is based on Prof. Konrad Jacobs’ (Erlangen) large collection of photographs of mathematicians from all over the world. In the 1950’s Prof. Jacobs started to make copies of the photographs he had taken and donated these copies to the MFO. In 2005 he transferred his whole collection completely with all rights to the MFO. The collection also contains a lot of pictures taken by Prof. George M. Bergman (Berkeley), whose pictures were scanned by the MFO and added to the database in 2010/11. Copyright Information / Disclaimer The Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO) owns the copyright to most of the images used on the Oberwolfach Photo Collection website. Those images labelled with "Copyright: MFO" can be used on the terms of the Creative Commons License Attribution-Share Alike 2.0 Germany. Chapter 11 Quotient Spaces Introduction We have seen how to create new topological spaces from given topological spaces using the operation of forming a subspace (of a topological space) and the operation of forming a (ﬁnite or inﬁnite) product of a set of topological spaces. In this chapter we introduce a third operation, namely that of forming a quotient space (of a topological space). As examples we shall see the Klein bottle and Möbius strip. 11.1 Quotient Spaces Let (X, τ ) and (Y, τ 1) be topological spaces. Then 11.1.1 Deﬁnitions. (Y, τ 1) is said to be a quotient space of (X, τ ) if there exists a surjective mapping f : (X, τ ) → (Y, τ 1) with the property (). For each subset U of Y , U ∈ τ 1 ⇐⇒ f −1(U ) ∈ τ . () A surjective mapping f with the property() is said to be a quotient mapping. 11.1.2 Remark. is a continuous map. From Deﬁnitions 11.1.1 it is clear that every quotient mapping 338 11.1. QUOTIENT SPACES 339 11.1.3 Remark. Property () is equivalent to property (). For each subset A of Y , A is closed in (Y, τ 1) ⇐⇒ f −1(A) is closed in (X, τ ). () Let f be a one-to-one mapping of a topological space (X, τ ) 11.1.4 Remark. onto a topological space (Y, τ 1). Then f is a homeomorphism if and only if it is a quotient mapping. 11.1.5 Proposition. (X, τ ) onto a Hausdorﬀ space (Y, τ 1). Then f is a quotient mapping. Let f be a continuous mapping of a compact space If f −1(A) Proof. We shall use property() above. Let A be a subset of Y . is closed in (X, τ ), then by Proposition 7.2.4 it is compact. As f is continuous, by Proposition 7.2.1, f (f −1(A)) is a compact subspace of the Hausdorﬀ space (Y, τ 1), and therefore by Proposition 7.2.5 is closed in (Y, τ 1). As f is surjective, f (f −1(A)) = A, and we have that A is closed. Conversely, if A is closed in (Y, τ 1), then by continuity of f , f −1(A) is closed in (X, τ ). Hence f has property(*) and so is a quotient mapping. As an immediate consequence of Proposition 11.1.5 and Remark 11.1.4 we obtain: Let f be a one-to one continuous mapping of a compact 11.1.6 Corollary. space (X, τ ) onto a Hausdorﬀ space (Y, τ 1). Then f is a homeomorphism. 340 CHAPTER 11. QUOTIENT SPACES Let (X, τ ) and (Y, τ 1) be topological spaces. A 11.1.7 Deﬁnitions. mapping f : (X, τ ) → (Y, τ 1) is said to be a closed mapping if for every closed subset A of (X, τ ), f (A) is closed in (Y, τ 1). A function f : (X, τ ) → (Y, τ 1) is said to be an open mapping if for every open subset A of (X, τ ), f (A) is open in (Y, τ 1). 11.1.8 Remark. mappings f which are In Exercises 7.2 #5 we saw that there are examples of (i) open but not closed; (ii) closed but not open; (iii) open but not continuous; (iv) closed but not continuous; (v) continuous but not open; (vi) continuous but not closed, and that if (X, τ ) and (Y, τ 1) are compact Hausdorﬀ spaces and f : (X, τ ) → (Y, τ 1) is a continuous mapping, then f is a closed mapping. Clearly if f : (X, τ ) → (Y, τ 1) is a surjective, continuous 11.1.9 Remark. and open mapping, then it is also a quotient mapping. Similarly if f is a surjective, continuous and closed mapping, then it is a quotient mapping. However, there exist quotient maps which are neither open nor closed maps. 11.1.10 Proposition. onto a Hausdorﬀ space (Y, τ 1). Then f is a closed mapping. Let f be a continuous mapping of a compact space Let A be a closed subset of (X, τ ). Then A is compact by Proposition Proof. 7.2.4. So by Proposition 7.2.1, f (A) is a compact subset of (Y, τ 1). As (Y, τ 1) is Hausdorﬀ, by Proposition 7.2.5, f (A) is a closed set. So f is a closed mapping. 11.1. QUOTIENT SPACES 341 Let f : [0, 1] → S1 be given by f (x) − (cos 2πx, sin 2πx). 11.1.11 Example. Clearly f is continuous and surjective. By Proposition 11.1.10, f is a closed mapping and by Remark 11.1.9 is therefore also a quotient mapping. Exercises 11.1 1. Verify Remark 11.1.3. 2. Verify Remark 11.1.4. 3. Verify both assertions in Remark 11.1.9. 4. Let (X, τ ), (Y, τ 1) and (Z, τ 2) be topological spaces and f : (X, τ ) → (Y, τ 1) and g : (Y, τ 1) → (Z, τ 2) be quotient maps. Prove that the map g ◦ f : (X, τ ) → (Z, τ 2) is a quotient map. (So the composition of two quotient maps is a quotient
map.) 5. Noting the deﬁnition of S1 in Exercises 6.1 #15, let f : R → S1 be given by f (x) = (cos 2πx, sin 2πx). Show that f is a quotient mapping but not a closed mapping. Is f an open mapping? 6. Find an example of a quotient mapping which is neither an open mapping nor a closed mapping. 7. Show that every compact Hausdorﬀ space is a quotient space of the Stone-˘Cech compactiﬁcation of a discrete space. 8. Let (X, τ ) and (Y, τ 1) be topological spaces and f : (X, τ ) → (Y, τ 1) a quotient mapping. Prove that if (X, τ ) is a sequential space then (Y, τ 1) is also a sequential space. (See Exercises 6.2.) 9. Let (X, τ ) and (Y, τ 1) be topological spaces and f : (X, τ ) → (Y, τ 1) a quotient mapping. If (X, τ ) is metrizable, is (Y, τ 1) necessarily metrizable? 10. Is a Hausdorﬀ quotient space of a kω-space necessarily a kω-space? 11. Is a quotient space of a k-space necessarily a k-space? 342 CHAPTER 11. QUOTIENT SPACES 12. Let (X, τ ) and (Y, τ 1) be topological spaces and f a quotient mapping of (X, τ ) onto (Y, τ 1). (i) If (X, τ ) is a metrizable space, prove that (Y, τ 1) is a sequential space. (See Exercises 6.2 #11.) (ii) If (X, τ ) is a ﬁrst countable space, prove that (Y, τ 1) is a sequential space. (iii)∗ Can sequential spaces be characterized as quotient spaces of metrizable spaces? 11.2 Identiﬁcation Spaces We begin by recalling the deﬁnition of an equivalence relation. 11.2.1 Deﬁnitions. A binary relation ∼ on a set X is said to be an equivalence relation if it is reﬂexive, symmetric and transitive; that is, for all a, b, c ∈ X, (i) a ∼ a (reﬂexive); (ii) a ∼ b =⇒ b ∼ a (symmetric); (iii) a ∼ b and b ∼ c =⇒ a ∼ c (transitive). If a, b ∈ X, then a and b are said to be in the same equivalence class if a ∼ b. Note that = is an equivalence relation on the set R of all real 11.2.2 Remark. numbers and ∼= (homeomorphic) is an equivalence relation on any set of topological spaces. 11.2.3 Remark. deﬁne an equivalence relation ∼ on the set X as follows: Let X and Y be sets and f a mapping of X onto Y . We can for a, b ∈ X, a ∼ b ⇐⇒ f (a) = f (b). So two points a, b ∈ X are in the same equivalence class if and only if f (a) = f (b). 11.2. IDENTIFICATION SPACES 343 11.2.4 Remark. Using Remark 11.2.3 we now observe that quotient spaces arise in a very natural way. Let (X, τ ) be any topological space and ∼ any equivalence relation on X. Let Y be the set of all equivalence classes of ∼. We can denote Y by X/ ∼. The natural topology to put on the set Y = X/ ∼ is the quotient topology under the map which identiﬁes the equivalence classes; that is, maps each equivalence class to a point. Because of this example, quotient mappings are often called identiﬁcation mappings and quotient spaces are often called identiﬁcation spaces. Of course, as we have seen in Remark 11.2.3, this example is in fact quite general, because if f is any mapping of a set X onto a set Y , we can deﬁne an equivalence relation by putting a ∼ b if and only if f (a) = f (b), where a, b ∈ X. For any space (X, τ ), the cone (CX, τ 1) over X is 11.2.5 Deﬁnition. the quotient space ((X, τ ) × I)/ ∼, where I denotes the unit interval with its usual topology and ∼ is the equivalence relation (x, 0) ∼ (x, 0), for all x, x ∈ X; that is, CX = ((X, τ ) × I)/(X × {0}). Intuitively1, CX is obtained from X × I by pinching X × {0} 11.2.6 Remark. to a single point. So we make X into a cylinder and collapse one end to a point. The elements of CX are denoted by x, a. It is readily veriﬁed that x → x, 1 is a homeomorphism of (X, τ ) onto its image in (CX, τ 1); that is, it is an embedding. So we identify (X, τ ) with the subspace {x, 1 : x ∈ X} ⊂ (CX, τ 1). 1A useful introductory book on algebraic topology covering topics such as cone and suspension is “Algebraic Topology” by Allen Hatcher and is freely downloadable from http://www.math.cornell.edu/~hatcher/AT/AT.pdf. Our presentation here is based on Brown [62]. 344 CHAPTER 11. QUOTIENT SPACES If in Remark 11.2.6 the space X is a circle in the euclidean 11.2.7 Example. space R2, then the cylinder X × I is a subspace of R3 and the cone CX is also a subspace of R3, as indicated in the diagrams below. Note that if X is a disk in R2 rather than a circle, then the cone CX is a solid cone. We need a little notation before discussing cones further. Let \| \| be the Euclidean norm on Rn deﬁned by 11.2.8 Deﬁnitions. \| x1, x2, . . . , xn \| = The cell En is deﬁned by En = {x ∈ Rn : \|x\| 1}. The ball Bn is deﬁned by Bn = {x ∈ Rn : \|x\| < 1}. The sphere Sn−1 is deﬁned by Sn−1 = {x ∈ Rn : \|x\| = 1}. 2 + . . . xn 2 + x2 x1 2. We see that S1 is the unit circle in R2, and E0 = B0 = {0} and S0 = {−1, 1}. Note that Sn−1 and En are closed bounded subsets of Rn, and so by the Generalized Heine Borel Theorem 8.3.3, both are compact. 11.2.9 Proposition. For m, n ∈ N, Em × En is homeomorphic to Em+n. Proof. Exercise. 11.2. IDENTIFICATION SPACES 345 11.2.10 Proposition. For each n ∈ N, CSn−1 is homeomorphic to En. Consider the map f : Sn−1 × I → En, given by f (s, x) = sx. (By xs Proof. we mean simply x times s.) From the deﬁnitions of Sn−1, I and En, clearly f is surjective and continuous. By Tychonoﬀ ’s Theorem 8.3.1 Sn−1 × I is compact, and so Proposition 11.1.10 implies f is a quotient mapping. As f −1{0} = Sn−1 × {0}, En−1 is homeomorphic to CSn. For any topological space (X, τ ) the suspension, 11.2.11 Deﬁnition. (SX, τ 2), is the quotient space of the cone (CX, τ 1) obtained by identifying the points X × {1}; that is, SX = CX/(X × {1}) = ([(X, τ ) × I]/[X × {0}])/(X × {1}). . Thus (SX, τ 2) is the quotient space of (X, τ ) × I where the equivalence relation is (x, 1) ∼ (x, 1) and (x, 0) ∼ (x, 0), for all x, x ∈ X. Intuitively, SX is obtained from X × I by pinching each of the sets X × {0} and X × {1}. If the space (X, τ ) is a circle, then the diagram below is a representation of SX. 346 CHAPTER 11. QUOTIENT SPACES 11.2.12 Proposition. of Sn−1 is homeomorphic to Sn. For each integer n 2, the suspension, S(Sn−1), Deﬁne the sets En + and En Proof. − as follows: En + = {(x, t) ∈ Sn : x ∈ Rn, t 0} = the northern hemisphere of Sn. − = {(x, t) ∈ Sn : x ∈ Rn, t 0} = the southern hemisphere of Sn. En Let p : Rn+1 = Rn ×R → Rn be the projection map which omits the last coordinate. Then consider the map p restricted to En + . Then p : En by the compactness of En + → En is continuous, one-to-one and onto and hence is a homeomorphism, +. Similarly p : En − → En is a homeomorphism. Now let φ : (X, τ ) × I → CX be the canonical quotient map, for any (X, τ ). and C−X = φ(X × [0, 1/2]. Put C+X = φ(X × [1/2, 1] It is readily veriﬁed that φ : (X, τ ) × [1 2, 1] → C+X and φ : (X, τ ) × [0, 1 2] → C−X are quotient maps. So we have a quotient map: −→ −→ −→ (X, τ ) × I (x, t) (X, τ ) × [0, 1/2] (x, t/2) which shrinks (X, τ ) × {0}. So CX is homeomorphic to C−X. Similarly the map I → [ 1 2 , yields CX is homeomorphic to C+X. C−X φ(x, t/2) 2, 1] given by t → 1 − t Putting our results together we have ∼= C+Sn−1 and En − En + − = {(x, 0) : x ∈ Rn}. They both map (x, 0) to (x, 1 En + ∩ En “glue” the homeomorphisms together to obtain Sn ∼= S(Sn−1). +. So ∼= C−Sn−1 and these homeomorphisms agree on the set 2).) Hence we can C+Sn−1 ∼= C+Sn−1 ∼= En ∼= En −→ 11.3. MÖBIUS STRIP, KLEIN BOTTLE AND REAL PROJECTIVE SPACE 347 Exercises 11.2 1. If (X, τ ) is a point, verify that the cone (CX, τ 1) is the unit interval [0, 1]. 2. Prove that every topological space (X, τ ) is homeomorphic to a subspace of a path-connected space by verifying that the cone CX is path-connected. 3. Prove Proposition 11.2.9. 4. Prove that En + is a retract of Sn, for n ∈ N. Reduced Cone and Reduced Suspension 5. Let (X, τ ) be a topological space and x0 ∈ X. Deﬁne the reduced cone, denoted by ΓX, as ΓX = (X × I)/(X × {0} ∪ {x0} × I) and the reduced suspension, denoted by ΣX, as ΣX = (X × I)/(X × {0, 1} ∪ {x0} × I). Prove that ΓSn−1 ∼= En and that ΣSn−1 ∼= Sn, for any positive integer n 2. 6. Let (X, τ ) be the topological space obtained from the real line R with the euclidean topology by indentifying the set of all integers to a point; that is n ∼ m if and only if n, m ∈ Z. Prove that (X, τ ) is a sequential space which is not a ﬁrst countable space. (See Exercises 6.2 #11.) 11.3 Möbius Strip, Klein Bottle and Real Projective Space 11.3.1 Remark. It is not always easy to picture a 3-dimensional object, let alone a 4-dimensional one which cannot exist in 3-dimensional space, such as the Klein bottle. It is therefore useful and convenient to use 2-dimensional polygonal representations of ﬁgures in higher dimensions, where we adopt a convention for when edges are identiﬁed. A simple example will demonstrate how this is done. 348 CHAPTER 11. QUOTIENT SPACES 11.3.2 Example. We represent a cylinder as a square with one pair of opposite sides identiﬁed. So the two sides labelled “b” are identiﬁed in such a way that the two vertices marked “A” are identiﬁed and the two vertices marked “B” are identiﬁed. So the cylinder is the quotient space I × I/ ∼, where ∼ is the equivalence relation on I × I given by (t, 0) ∼ (t, 1), for all t ∈ I. 11.3. MÖBIUS STRIP, KLEIN BOTTLE AND REAL PROJECTIVE SPACE 349 11.3.3 Example. is deﬁned to be the quotient space I × I/ ∼, where ∼ is the equivalence relation (t, 0) ∼ (1 − t, 1), for all t ∈ I. The Möbius strip or Möbius band, denoted by M, If an insect crawled along the entire length of the Möbius strip, it would return to its starting point without ever crossing an edge, but it would have crawled along both sides of the Möbius strip. 350 CHAPTER 11. QUOTIENT SPACES The real projective plane, denoted by RP2, is deﬁned to 11.3.4 Examples. be the quotient space I × I/ ∼, where ∼ is the equivalence relation (0, t) ∼ (1, 1 − t) and (t, 0) ∼ (1 − t, 1), for all t ∈ I. This is not a subspace of R3, that is, it cannot be embedded in 3-dimensional euclidean space (without crossing itself ). While it is not immediately obvious, RP2 can be thought of as the space of all lines
through the origin, but excuding the origin, in R3. Each line is of course determined by a non-zero vector in R3, unique up to scalar multiplication. RP2 is then the quotient space of R3 \ {0} under the equivalence relation v ∼ λv, for all λ ∈ R, λ = 0. (See also Deﬁnitions A5.0.1 and the following discussion.) In fact, we can generalize this to real projective space, denoted by RPn. Let X = Rn+1 \ {(0, 0, . . . , 0)}. Give X the topology τ as a subspace of Rn+1. Then RPn is the quotient space (X, τ )/∼, where the equivalence relation ∼ is given by (a1, a2, . . . , an+1) ∼ (b1, b2, . . . , bn+1), if there exists a λ ∈ R \ {0}, such that bi = λai, for i = 1, 2, . . . , n + 1. Let φ : (X, τ ) → RPn be the quotient/identiﬁcation map. Since Sn ⊆ X, we can consider the restriction of φ to Sn, namely φ : Sn → RPn. Since φ((x1, x2, . . . , xn+1)) = φ(( x1, 1 λ xn+1)), 1 λ x2, . . . , 1 λ n+1 and ( 1 where λ = \|(x1, x2, . . . , xn+1)\| = Sn, we see that the map φ : Sn → RPn is surjective. It is continuous and, by the compactness of Sn, is a quotient map. Clearly if a, b ∈ Sn, then λxn+1) ∈ λx1, 1 λx2, . . . , 1 2 + · · · + x2 1 + x2 x2 φ(a) = φ(b) ⇐⇒ a = (a1, a2, . . . , an+1) and b = (−a1, −a2, . . . , −an+1); that is, a and b are antipodal points. (Antipodal points on a sphere are those which are diametrically opposite.) So RPn ∼= Sn/ ≈, where ≈ is the equivalence relation on Sn which identiﬁes antipodal points. 11.3. MÖBIUS STRIP, KLEIN BOTTLE AND REAL PROJECTIVE SPACE 351 The quotient space I × I/ ∼, where ∼ is the equivalence 11.3.5 Example. relation (0, t) ∼ (1, t) and (t, 0) ∼ (1 − t, 1), is called the Klein bottle2, denoted by K. Like the real projective plane in Examples 11.3.4, the Klein bottle cannot be embedded in 3-dimensional euclidean space (without crossing itself ), but it can be embedded in R4. When embedded in R4, like the Möbius strip, it is one-sided. Exercises 11.3 1. Verify that the Möbius strip is a compact Hausdorﬀ subspace of R3 with boundary homeomorphic to a circle. 2The beautiful representation on this page of the Klein bottle in 3-dimensions was produced by Professor Thomas F. Banchoﬀ, a geometer, of Brown University, Providence, Rhode Island, USA. 352 CHAPTER 11. QUOTIENT SPACES 2. The diagram below represents the quotient space I × I/∼. Write down the equivalence relation ∼ and then verify that the diagram represents a torus. Deduce that the torus is a compact Hausdorﬀ topological space. 3. Verify that the boundary of the Klein bottle in R4 is the empty set. 11.4 Postscript In this chapter we introduced the important notions of quotient space and its associated quotient mapping or identiﬁcation mapping. We noted that every continuous mapping of a compact space onto a Hausdorﬀ space is a quotient mapping. Quotients were used to produce new and interesting spaces including cylinders, cones, the Klein bottle, real projective spaces and the Möbius strip. We introduced the cone and suspension which are of relevance to study in algebraic topology. In the ﬁnal section we showed how to use polygonal representations to deﬁne ﬁgures such as the Klein bottle which cannot be embedded in 3-dimensional euclidean space. This short chapter is but the smallest taste of what awaits you in further study of topology. 11.5. CREDIT FOR IMAGES 11.5 Credit for Images 1. Klein Bottle. Dear Professor Morris, 353 You have my permission to use our Klein bottle in your freely available online book "Topology Without Tears". The parametrization was due to me and the particularly nice rendering was produced by a student, Jeﬀ Beall. Of all the images that have come from our work, this is the one most requested for reproduction. Tom Banchoﬀ, Professor Emeritus Brown University (as of 2014) August 8, 2017. Chapter 12 The Stone-Weierstrass Theorem Introduction Section 1 is devoted to the Weierstrass Theorem, proved by Karl Weierstrass (1815–1897) in 1885 when he was aged 70. It is a very powerful theorem, with very many applications even to this day. Over the next 30 years there was a variety of proofs by Weierstrass Stone famous mathematicians. In 1964 Kuhn [245], gave an elementary proof using the Bernoulli inequality (named after Jacob Bernoulli (1655–1705)). In Section 1 we give a quite elementary1 proof using the Bernoulli inequality. The proof we present here is primarily due to my teacher, Rudolf Výborný in his paper Výborný [414] but is also inﬂuenced by the presentation of Păltineanu [333] which was also inﬂuenced by Výborný [414]. Bernoulli Výborný In Section 2, we prove the generalization of the Weierstrass Theorem ﬁrst proved in 1937 by Marshall Harvey Stone (1903–1989), Stone [384], and with a simpler proof in 1948, Stone [385]. This elegant generalization is known as the Stone-Weierstrass Theorem. The beautiful Stone-Weierstrass Theorem is a theorem in Pure Mathematics but has very important applications. 1Readers should understand that elementary does not mean easy. Rather it means that it uses only elementary mathematics. 354 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 355 12.1 The Weierstrass Approximation Theorem We begin by stating and proving the Bernoulli inequality which was named after the Swiss mathematician Jacob Bernoulli (1654–1705). 12.1.1 Proposition. Let x ∈ R such that x −1, and n ∈ N. Then (1 + x)n 1 + nx. Proof. This is proved using mathematical induction. Clearly it is true if n = 1. Assume that for some integer k, (1 + x)k 1 + kx, for all x −1. Then (1 + x)k+1 = (1 + x)k.(1 + x) (1 + kx).(1 + x), = 1 + kx + kx + kx2 1 + kx + kx by the inductive assumption = 1 + (k + 1)x, as required. So if the Bernoulli inequality is true for n = k, then it is true for n = k + 1. Thus by mathematical induction, the Bernoulli inequality is true for all n ∈ N. We shall prove the Weierstrass Theorem on Polynomial Approximation. [a, b] → C is a The Weierstrass Approximation Theorem. continuous function and ε is a positive real number, then there exists a polynomial P such that \|f (x) − P (x)\| < ε, for all x ∈ [a, b]. If f : 356 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM If you think about this, you can see that it is quite surprising. It says that no matter how strange a continuous function may be, it can be approximated as close as we like by a very nice simple function, namely a polynomial2. Not only is this result quite surprising, it is very useful in practical applications. We shall prove the Weierstrass Approxmation Theorem with C replaced by R, from which the Theorem easily follows. The main idea of the proof is to show that if f can be approximated by a polynomial on some interval, then it can be approximated by a polynomial on a slightly larger interval. To show this we shall use the following lemma. First we mention a deﬁnition: Let an ∈ C and b ∈ C, for each n ∈ N. 12.1.2 Deﬁnition. If for each ε > 0, there exists an Nε ∈ N such that \|an − b\| < ε, for n Nε, then the sequence of complex numbers an is said to have b as its limit. This is denoted by lim n→∞ an = b or equivalently an → b. 2Even if the function f is continuous but nowhere diﬀerentiable, it can be approximated by an inﬁnitely diﬀerentiable function, namely a polynomial. This remarkable result is not to be confused with Taylor polynomials or Taylor series which require f to be not just diﬀerentiable but inﬁnitely diﬀerentiable. See Exercises 10.3 #28(vii). 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 357 Let X be an arbitrary set, U a function mapping X into 12.1.3 Lemma. R such that 0 < U (x) < 1, for all x ∈ X. Let A and B be disjoint subsets of X. If there exists a k ∈ N such that U (x) < 1 , for k all x ∈ B, then , for all x ∈ A and U (x) > 1 k (1 − [U (x)]n)kn (1 − [U (x)]n)kn lim n→∞ lim n→∞ = 1, for all x ∈ A, and = 0, for all x ∈ B. Proof. 1 1 − [U (x)]n)kn 1 − kn[U (x)]n, by Proposition 12.1.1 = 1 − [kU (x)]n → 1, for x ∈ A. 0 1 − [U (x)]n)kn [kU (x)]n (1 + kn[U (x)]n) (1 + [U (x)]n)kn [kU (x)]n , by Proposition 12.1.1 = (1 − [U (x)]n)kn [kU (x)]n (1 − [U (x)]n)kn < [kU (x)]n (1 − [U (x)]n)kn (1 − [U (x)]2n)kn [kU (x)]n 1 [kU (x)]n = < → 0, for x ∈ B. 358 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM Let a, b, c, d ∈ R with a < d < c < b and let η be any 12.1.4 Lemma. postive real number. Then for each n ∈ N, there exists a polynomial pn such that 0 pn(x) 1, 1 − η pn(x), pn(x) η, pn(x) = 1, for x ∈ [a, b]; for x ∈ [a, d]; for x ∈ [c, b]; for x ∈ [a, d]; pn(x) = 0, for x ∈ [c, b]. lim n→∞ lim n→∞ (1) (2) (3) (4) (5) Proof. Here is an example Let e = U (x) = c + d 2 , and 1 2 + x − e 2(b − a) , for all x ∈ [a, b]. 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 359 Noting that a < d < e < c < b we have U (a) = U (d) = U (c) = U (b(b − a) d − e 2(b − a) c − e 2(b − a) b − e 2(b − a) b − a 2(b − a) a − e 2(b − a(b − a) From its deﬁnition we see that U is an increasing function and it then follows from the above inequalities that U (x) < 0 < U (xx) > , , x ∈ [a, b] x ∈ [a, d] x ∈ [c, b] Deﬁne the polynomial pn, for each n ∈ N, by pn(x) = 1 − [U (x)]n2n , for all x ∈ [a, b]. Clearly we have 0 pn(x) 1 , for all x ∈ [a, b] which proves (1) in the statement of the Lemma. Applying Lemma 12.1.3 for k = 2, A = [a, d], and B = [c, b], we have that lim n→∞ lim n→∞ pn(x) = 1, for x ∈ [a, c] ; pn(x) = 0, for x ∈ [d, b] , 360 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM which proves (4) and (5) in the statement of the Lemma. By (4) there exists n1 ∈ N such that pn1(x) > 1 − η, for all x ∈ [a, d] and by (5) there exists n2 ∈ N such that pn2(x) < η, for all x ∈ [c, b]. Putting n equal to the greater of n1 and n2, we have that (4) and (5) imply (2) and (3), which completes the proof of the Lemma. 12.1.5 Theorem. [The Weierstrass ApproximationTheorem] If f : [a, b] → C is a continuous function and ε is a positive real number, then there exists a polynomial P such that \|f (x) − P (x)\| < ε, for all x ∈ [a, b]. (6) Let g(x) = (f (x)) and h(x) = (f (x)) the real and imaginary parts of Proof. f (x) respectively, for x ∈ [a, b], so that g and h are continuous functio
ns from [a, b] into R. If there exist polynomials Pg and Ph such that for all x ∈ [a, b], \|g(x)−Pg(x)\| < ε 2 2, then putting P (x) = Pg(x) + iPh(x) we have that the and \|h(x) − Ph(x)\| < ε polynomial P satisﬁes \|f (x) − P (x)\| < ε. So to prove this theorem, it suﬃces to prove it for the special case: f is a continuous function of [a, b] into R. We shall now assume, without loss of generality, that f is indeed a continuous function of [a, b] into R. For the given ε > 0, let Sε be the set of all t b such that there exists a polynomial Pε with the property that \|f (x) − Pε(x)\| < ε, for all x ∈ [a, t]. We need to show Sε is not the empty set. By continuity of f , there exists t0 > a such that \|f (x) − f (a)\| < ε, for all x ∈ [a, t0]. (7) (8) 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 361 Consequently, f can be approximated by the constant function f (x) = f (a) on [a, t0], so Sε = Ø. Let s = sup Sε. Clearly a < s b. The proof of the Theorem will be complete if we can prove that s = b. By continuity of f at s, there is a δ > 0 such that \|f (x) − f (s)\| ε 3 , for s − δ x s + δ and a x b. (9) By the deﬁnition of supremum, there is a c with s − δ c s and c ∈ Sε. This means that there is a polynomial Pε satisfying (7) for x ∈ [a, c]. Let m = max{\|f (x) − Pε(x)\| : a x c}. So m < ε. (10) Recall Proposition 7.2.15 says that if φ is a continuous function from [a, b] into R, then φ([a, b]) = [v, w], for some v, w ∈ R. As the function f and the polynomial Pε are continuous on [a, b], the function φ deﬁned by φ(x) = \|f (x) − Pε(x)\| + \|f (x) − f (s)\| is continuous on [a, b]. So φ([a, b]) = [v, w], for some v, w ∈ R. Therefore we can choose M ∈ R such that M > \|f (x) − Pε(x)\| + \|f (x) − f (s)\|, for all x ∈ [a, b] (11). We shall apply Lemma 12.1.4 for d = s − δ. Noting that, by (10), m < ε, choose 0 < η < 1 so small that m + M η < ε and M η < 2ε 3 . Apply Lemma 12.1.4 to ﬁnd a polynomial pn satisfying (1), (2) and (3). Now we deﬁne the required polynomial P by P (x) = f (s) + \|Pε(x) − f (s)]pn(x). (12) (13) and we shall show that it satisﬁes condition (6) of the statement of the Theorem. 362 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM First we have \|f (x) − P (x)\| = = = f (x) − f (s) − [Pε(x) − f (s)]pn(x) [f (x) − Pε(x)]pn(x) + [f (x) − f (x)pn(x) − f (s) + f (s)pn(x)] [f (x) − Pε(x)]pn(x) + [(f (x) − f (s))(1 − pn(x)] pn(x) + \|f (x) − f (s) f (x) − Pε(x) (1 − pn(x)) (14) On the interval [a, s − δ] \|f (x) − P (x)\| \|f (x) − Pε(x)\| − \|f (x) − Pε(x)\|(1 − pn(x)) + \|f (x) − f (s)\|(1 − pn(x)) , by (14) \|f (x) − Pε(x)\| + \|f (x) − Pε(x)\| + \|f (x) − f (s)\|(1 − pn(x)) m + M η , by (10), (11) and (2) as [a, s − δ] ⊆ [a, c] < ε , by (12). (15) On the interval [s − δ, c] \|f (x) − P (x)\| f (x) − Pε(x) pn(x) + \|f (x) − f (s) (1 − pn(x)), by (14) εpn(x) + ε 3 (1 − pn(x)), by (10) and (9) noting that , by (1). (16) On the interval [c, s + δ] ∩ [c, b] we have \|f (x) − P (x)\| f (x) − Pε(x) pn(x) + \|f (x) − f (s) (1 − pn(x)), by (11) and (9) as s − δ c s b (1 − pn(x)), by (14) ε 3 M pn(x) + 2ε 3 ε 3 = ε , by (1). < + , by (2), (1), and the second part of (16) (17) By (15),(16), and (17), we have \|f (x) − P (x)\| < ε, for all x ∈ [a, s + δ] ∩ [c, b]. (18) Suppose s < b. Then (18) contradicts the deﬁnition of s as the supremum of the set Sε. So s = b; that is, [a, s] = [a, b], which completes the proof of the Theorem. 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 363 12.1.6 Remark. In the literature there are many somewhat diﬀerent proofs of the Weierstrass Approximation Theorem 12.1.5. Our proof above, uses proof by contradiction which is avoided in the constructive proof using Bernstein Polynomials, discovered in 1912 by the Russian mathematician Sergei Natanovich Bernstein (1880–1968), and which is motivated by probability theory. Bernstein [39] proved that for any continuous function f : [0, 1] → R, f can be expressed as an inﬁnite sum of polynomials: Bernstein f (x) = lim n→∞ n m=0 f m n m n xm(1 − x)n−m, and so f can be approximated as closely as we like by the partial sum from m = 0 xm(1 − x)n−m is to m = n. For each n ∈ N, the polynomial Bn(=0 called the Bernstein polynomial of degree n associated with f . (See Bernstein basis polynomials in Exercises 10.3 #28.) So Bernstein polynomials provide a very concrete and practical method of approximating continuous functions. We will see this is extemely useful in Remark 12,1.7. [0, 1] → R with f m n Noting that for any a0, a1, a2, . . . , an ∈ R, there exists a continuous function = am, m = 0, 1, 2, . . . , n (Exrercises 12.1 #5) we f : n xm(1 − x)n−m is the Bernstein polynomial m=0 of degree nassociated with some continuous function f : [0, 1] → R. But, noting Exercises 10.3 #28 (ix), we see that every polynomial can be written in this form. So see that every polynomial m n am every polynomial is the Bernstein polynomial associated with some continuous function f : [0, 1] → R. 364 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12.1.7 Remark. It is worth mentioning that Bernstein polynomials are at the heart of what are known as Bézier curves used in graphics. They were developed by the French mathematician Paul de Casteljau (1930–) at Citroën and the French engineer Pierre Bézier (1910–1999) at Renault who used them to design car bodies. Today this mathematics is at the core of computer graphics and CAD/CAM (computer-aided-design/computer- Casteljau aided manufacturing). Quoting from “The ﬁrst years of CAD/CAM and the UNISURF CAD System” by Pierre Bézier, in Piegl [325]: In the words of Pierre Bézier himself: There is no doubt that Citroën was the ﬁrst company in France that paid attention to CAD, as early as 1958. Paul de Casteljau, a highly gifted mathematician, devised a system based on the use of Berstein polynomials . . . the system devised by de Casteljau was oriented towards translating already existing shapes into . . . patches, deﬁned in terms of numerical data. Bezier Due to Citroën’s policy, the results obtained by de Casteljau were not published until 1974, and this excellent mathematician was deprived of part of the well deserved fame that his discoveries and inventions should have earned him. We can deduce a stronger version of The Weierstrass Approximation Theorem 12.1.6 as a corollary of the theorem itself. If P is such that P (x) = a0+a1x+. . . anxn, where each aj = rj + isj, where aj and bj are rational (real) numbers for j ∈ {1, 2, . . . , n}, then the polynomial P is said to have rational number coeﬃcients. If f : [a, b] → C is a continuous function and ε is a 12.1.8 Corollary. positive real number, then there exists a polynomial P with rational number coeﬃcients such that \|f (x) − P (x)\| < ε, for all x ∈ [a, b]. Proof. Exercise. 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 365 Let C[0, 1] be the set of all continuous functions of 12.1.9 Proposition. [0, 1] into R. If τ is the topology on C[0, 1] induced by the supremum metric in Example 6.1.6, then (i) the set of all polynomials is dense in (C[0, 1], τ ); (ii) the set of all polynomials with rational number coeﬃcients is dense in (C[0, 1], τ ); (iii) (C[0, 1], τ ) is a separable space; and (iv) the cardinality of C[0, 1] is 2ℵ0. Proof. Exercise. To make the proof of The Weierstrass Approximation Theorem 12.1.5 as elementary as we could, we avoided any mention of uniform convergence. But we do mention it now before completing this section. Let S be any subset of R and pn, n ∈ N, a sequence 12.1.10 Deﬁnition. of functions S → C. Then pn, n ∈ N is said to be pointwise convergent to a function f : S → C if given any x ∈ S and any ε > 0, there exists N (x, ε) ∈ N such that \|f (x) − pn(x)\| < ε, for every n ∈ N such that n > N (x, ε). Note that the notation N (x, ε) means that this number depends on both x and ε. Contrast this deﬁnition with the next one. Let S be any subset of R and pn, n ∈ N, a sequence 12.1.11 Deﬁnition. of functions S → C. Then pn, n ∈ N, is said to be uniformly convergent to a function f : S → C if given any ε > 0, there exists N (ε) ∈ N such that \|f (x) − pn(x)\| < ε, for every x ∈ S and n ∈ N such that n > N (ε). 366 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12.1.12 Remark. Clearly uniform convergence of a sequence of functions implies pointwise convergence of that sequence. However, the converse is false. For example, it is easy to see that if pn(x) = nx 1+n2x2 , for all x ∈ S = (0, ∞), then pn, n ∈ N, is pointwise convergent to the function f (x) = 0, for all x ∈ (0, ∞). n) n) − pn( 1 However, for every 0 < ε < 1 2 > ε and so pn, n ∈ N, is not uniformly convergent to f . Thus pointwise convergence does not imply uniform convergence , 12.1.13 Remark. We can now restate The Weierstrass Approximation Theorem 12.1.5: If f : [a, b] → C is a continuous function, then there exists a sequence pn, n ∈ N, of polynomials on [a, b] which is uniformly convergent to f . 12.1.14 Remark. Having seen in the Weierstrass Approximation Theorem 12.1.5 that a continuous function can always be approximated by a polynomial, it is perhaps appropriate to underline the diﬀerence in behaviour of continuous functions and polynomials, indeed of (i) continuous functions, (ii) analytic functions (iii) C∞ functions and (iv) polynomials. we begin with some deﬁnitions. Let U be an open subset of R. A function f : U → R 12.1.15 Deﬁnitions. is said to be smooth if it is inﬁnitely diﬀerentiable at every point x0 ∈ U . The function f is said to be analytic at a point x0 ∈ U if there exists an open neighbourhood O ⊆ U of x0 such that f is inﬁnitely diﬀerentiable at every x ∈ O, and the Taylor series T (x) = ∞ n=0 f (n)(x0) n! (x − x0)n converges (pointwise) to f (x) for all x in O, where f (n)(x0) denotes the nth derivative of f evaluated at x0. The function f is said to be analytic on RRR if it is analytic at every x0 ∈ R. The set of analytic functions on R properly contains the set of all polynomials on R, but is a proper subset of the set C∞ of smooth (ie inﬁnitely diﬀerentiable) 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 367 functions on R. An example
of a smooth function which is not analytic is given in Exercises 12.1 #10. 12.1.16 Theorem. of zeros of f ; that is, Z = {x : x ∈ R such that f (x) = 0}. Let f be a function of R into itself and Z be the set (i) If f is a non-constant polynomial, then Z is a ﬁnite set. (ii) If f is a non-constant analytic function, then Z is a discrete subspace of R. (iii) If Z is any closed subset of R, then a C∞ function f can be chosen with Z its set of zeros. Proof. Exercise. Having discussed diﬀerentiability of a function f : R → R, 12.1.17 Remark. we conclude this section by discussing diﬀerentiability3 of a function f : Rn → Rm, a topic that is needed for the important study of diﬀerentiable manifolds. Recall that a function f : R → R is said to be diﬀerentiable at a point x0 ∈ R if there exists a number f (x0) ∈ R such that for h ∈ R f (x0 + h) − f (x0) h converges to f (x0) as h → 0. We usually write this as f (x0 + h) − f (x0) h lim h→0 = f (x0). (1) We reformulate this so that it can be generalized in a natural way to higher dimensional euclidean space. If we deﬁne a linear transformation λ : R → R by λ(h) = f (x0)h, then equation (1) becomes f (x0 + h) − (f (x0) + λ(h)) h lim h→0 = 0. (2) 3See Spivak [370]. 368 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM Let U be an open subset of Rn, n ∈ N. A function 12.1.18 Deﬁnition. f : U → Rm, m ∈ N, is said to be diﬀerentiable at a point x0 ∈ U , if there exists an open neighbourhood O ⊆ U of x0 such that for some linear transformation λ : Rn → Rm lim h→0 \|f (x0 + h) − (f (x0) + λ(h))\| \|h\| where h ∈ Rn and \|.\| denotes the euclidean norm both in Rn and Rm. The linear transformation λ is said to be the derivative of f at x0 and is denoted by Df (x0). = 0 12.1.19 Proposition. If f : U → Rm, m ∈ N, is diﬀerentiable at x0 ∈ U , then there is a unique linear transformation λ : Rn → Rm such that Let U be an open subset of Rn, n ∈ N. \|f (x0 + h) − (f (x0) + λ(h))\| \|h\| lim h→0 = 0 . Proof. Exercise. At this stage, the unique linear transformation in Proposition 12.1.19 and Deﬁnition 12.1.18 is somewhat mysterious. We proceed to remove that mystery. 12.1.20 Deﬁnition. and r = (r1, r2, . . . , rn) ∈ U , the limit Let U be an open subset of Rn, n ∈ N. If f : U → R f (r1, r2, . . . ri−1, ri + h, ri+1, . . . , rn) − f (r1, r2, . . . , rn) h lim h→0 if it exists is said to be the ith partial derivative of f at r and is denoted by Dif (r). 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 369 Of course if we put g(x) = f (r1, r2, . . . , ri−1, x, ri+1, . . . , rn), then Dif (r) equals g(r), the derivative of the function g at ri. Let U be an open subset of Rn, n ∈ N, and 12.1.21 Proposition. f : U → Rm, m ∈ N. Put f = (f 1, f 2, . . . , f m), where each f i : U → R. If f is diﬀerentiable at x0 ∈ U , then Djf i(x0) exists for 1 i m, 1 j n and f (x0) is the m × n matrix (Djf i(x0)). We now see that it is quite straightforward to calculate the derivative of a diﬀerentiable function mapping Rn into Rm. For a much more detailed discussion of the derivative of functions of several variables, see Spivak [370]. Exercises 12.1 1. Prove Corollary 12.1.8. 2. Prove Proposition 12.1.9. 3. Prove Lemma 12.1.4 with η replaced by 1 n. [Hint. Be careful, as this is not as trivial as it ﬁrst appears to be.] 4. Calculate the Bernstein polynomials B1(f ) and B2(f ) of each of the functions f = f1, f = f2, and f = f3, where f1(x) = x, x ∈ [0, 1], f2(x) = x2, x ∈ [0, 1], and f3(x) = x3, x ∈ [0, 1], respectively. 5. Let x0, x1, . . . , xn ∈ [0, 1], with x0 < x1 < · · · < xn, and let a0, a1, . . . , an ∈ R. Show that there exists a continuous function f : [0, 1] → R such that f (xi) = ai, for i = 0, 1, . . . , n. 6. Prove the following statement (Rudin [352], Theorem 7.8): Let pn, n ∈ N, be a sequence of functions on a subset S of R. The sequence pn, n ∈ N, is uniformly convergent to some function f : S → R if and only if for each ε > 0, there exists N ∈ N such that \|pm(x) − pn(x)\| < ε, for all x ∈ S and m, n ∈ N with m, n > N. 370 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 7. (i) Let S be a subset of R and pn, n ∈ N, a sequence of continuous functions of S into C. If the sequence pn, n ∈ N, is uniformly convergent to the function f : S → C, prove that f is a continuous function. [Rudin [352], Theorem 7.12.] (ii) Let C[0, 1] denote the normed vector space of all continuous functions \|f (x)\|. Using (i) prove that C[0, 1] is f : [0, 1] → R, where \|\|f \|\| = sup x∈[0,1] a Banach space. (iii) Let X be a compact Hausdorﬀ space and F be R or C. Prove that the normed vector space C(X, F ) of all continuous functions f : X → F with the sup (or uniform) norm, given by \|\|f \|\| = sup x∈X \|f (x)\|. is a Banach space. Dini’s Theorem 8. Dini’s Theorem. (Rudin [352], Theorem 7.13.) Let K be a subset of R and pn, n ∈ N, a sequence of continuous functions mapping K into R such that the sequence pn, n ∈ N, is pointwise convergent to a continuous function f : K → R. If (i) K is compact and (ii) pn(x) pn+1(x), for all x ∈ K, then the sequence pn, n ∈ N, is uniformly convergent to f . [Dini’s Theorem is named after the Italian mathematician Ulisse Dini (1845–1918).] 9. Verify Theorem 12.1.5 (i). Dini Support of a Function and Bump Functions 10. The support of a function f : R → R is {x : f (x) = 0}. A function f : R → R is said to be a bump function if it is smooth and its support is a compact subset of R. Verify that the function b : R → R given by b(x) = − 1 1−x2 exp 0 for \|x\| < 1 for \|x\| 1 is a bump function, and so is a smooth function, but is not an analytic function. 12.1. THE WEIERSTRASS APPROXIMATION THEOREM 371 Uniformly Convergent Series 11. Let Z be any closed subset of R. Verify the following: (i) R \ Z is the union of open intervals Ui, i ∈ I ⊆ N, where at most two of the Ui are unbounded. (ii) For each i ∈ I, let bi be a bump function on Ui such that if Ui is unbounded, then bi is constant outside some bounded interval. (See Exercise 10 above.) (x)\| < 2−i, for 0 j i, for all We can choose ci > 0 such that \|cib x ∈ R, where b (iii) Deﬁne f (x) = denotes the jth derivative of the bump function bi. i∈I cifi(x). Then each series i∈I cif (j)(x) converges (j) i (j) i absolutely and unformly for all x ∈ R. [Recall that a series ∞ n=1 θn(x), x ∈ R, is said to be uniformly convergent if for every ε > 0, there exists an N , (N independent of x), such that for all n N and all x ∈ R, \|sn(x) − s(x)\| < ε, where sn(x) = n k=1 θk(x) and s(x) = ∞ k=1 θk(x).] (iv) Then f is a smooth function. (v) So Theorem 12.1.16 (iii) is true. (vi) The function f : R → R is a smooth function but not an analytic function, where f (x) = − 1 1−x2 , e 0, if \|x\| < 1, otherwise. [Hint: Use Theorem 12.1.16 (ii).] 372 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12. Let f : R → R be a non-constant function analytic on R. Let Z = {x : f (x) = 0}. Assume that there exists an x0 ∈ R such that f (x0) = 0. As f is analytic, there exists an open neighbourhood O of x0 such that ∞ n=0 f (n)(x0) n! (x − x0)n → f (x), for all x ∈ O. Putting an = f (n)(x0) n! , this says ∞ n=0 an(x − x0)n → f (x), for all x ∈ O. Since f (x0) = 0, a0 = 0. Without loss of generality, assume a0 = a1 = · · · = ak = 0 and ak+1 = 0, where k 0. So we can write the Taylor series for f about x0 as ∞ an(x − x0)n = (x − x0)k ∞ an+k(x − x0)n = (x − x0)kg(x) n=0 n=k where g(x) = ∞ n=0 an+k(x − x0)n and is analytic in the open neighbourhood O of x0. So g is a continuous function of O into R. Since g(x0) = ak = 0, verify each of the following: (i) There exists an ε > 0 such that for \|x − x0\| < ε, \|g(x) − ak\| < \|ak\| 2 . (ii) g(x) = 0, for \|x − x0\| < ε. (iii) Z ∩ (x0 − ε, x0 + ε) = {x0}. (iv) From (iii) above, Z is a discrete countable subspace of R and so Theorem 12.1.15(ii) is true. 13. Prove Proposition 12.1.19. 14. Prove that if f : Rn → Rm, n, m ∈ N, is a constant function (that is, for some y ∈ Rm, f (x) = y for all x ∈ Rn), then Df (x) = 0, for all x ∈ Rn. 15. Prove that if f : Rn → Rm, m, n ∈ N, is a linear transformation, then for each x ∈ Rn, Df (x) = f (x). 12.2. THE STONE-WEIERSTRASS THEOREM 373 16. Prove that if f, g : Rn → R, n ∈ N, are diﬀerentiable at x0 ∈ Rn, then D(f + g)(x0) = Df (x0) + Dg(x0), D(f g(x0)) = g(x0)Df (x0) + f (x0)Dg(x0), and g(x0)Df (x0) − f (x0)Dg(x0) (g(x0))2 17. Find the partial derivatives of the following: D(f /g)(x0)) = , for g(x0) = 0. (i) f (x, y, z) = xy + z. (ii) f (x, y, z) = cos(xy) + sin(z). 18. Prove Proposition 12.1.21. 19. Let f : Rn → Rm, m, n ∈ \|N . Deﬁne what it should mean that f is a smooth function and what it should mean that f is an analytic function. 12.2 The Stone-Weierstrass Theorem In §1 we saw that the set of polynomials can be used to approximate any continuous function of [0, 1] into R, or put diﬀerently, the set of all polynomials is dense in C([0, 1], R) with the supremum metric. We immediately saw generalizations of this, namely that the smaller set of all Bernstein polynomials is dense in C([0, 1], R) as is the set of all polynomials with rational number coeﬃcients. So we might ask: which subsets of C([0, 1], R) are dense? But we shall see much more. The Stone-Weierstrass Theorem addresses the more general problem of identiﬁying the dense subsets of C(X, R) and C(X, C), where X is a compact Hausdorﬀ space. The Weierstrass Approximation Theorem 12.1.5 is a special case. To address this problem we shall introduce some new concepts and deﬁnitions which are of importance in their own right. First, we make an insightful observation. 374 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM Let S be a set of functions from a set X into a set Y . 12.2.1 Deﬁnition. Then S is said to separate points (of X) if for each a, b ∈ S with a = b, there exists a φ ∈ S such that φ(a) = φ(b). 12.2.2 Examples. (a) The set of all real-valued polynomials separates points of [0, 1] as for any a, b ∈ [0, 1] with a = b, the polynomial p given by p(x) = x, for all x ∈ [0, 1] is such that pa(a) = a = b = p(b). (b) The set of all real-valued Bernstein polynomials separates points of [0, 1] as
for any a, b ∈ [0, 1] with a = b, the Bernstein polynomial B1(f ), where f (x) = x, for all x ∈ [0, 1], is readily seen to satisfy B1(f ) = f . So (B1(f ))(a) = (B1(f ))(b). (c) On the other hand, the set {fn : fn(x) = sin(2πnx), x ∈ [0, 1], n ∈ N} of functions of [0, 1] into R does not separate the points 0 and 1, since fn(0) = sin(0) = 0 = sin(2πn) = fn(1), for all n ∈ N. Our next proposition gives us a necessary (but not suﬃcient) condition for a subset S of C[0, 1] to be dense, namely that it separates points of [0, 1]. Indeed it provides a necessary condition for a subset S of C(X, F ) to be dense, for F equal to R or C, and X a compact Hausdorﬀ space. 12.2. THE STONE-WEIERSTRASS THEOREM 375 Let (X, τ ) be a compact Hausdorﬀ space, F equal 12.2.3 Proposition. to R or C, and S a subset of C(X, F ), the set of all continuous function of (X, τ ) into F with the topology induced by the supremum metric. If S is dense in C(X, F ), then S separates points of X. Suppose S does not separate points of X. Then there exist a, b ∈ X with Proof. a = b such that φ(a) = φ(b), for all φ ∈ S. As S is dense in C(X, F ), for each ε > 0 and each f ∈ C(X, F ), there exists a φ ∈ S such that \|f (x) − φ(x)\| < ε. sup x∈X As X is compact Hausdorﬀ, there is a continuous function f ∈ C(X, F ) such that f (a) = 0 and f (b) = 1, and put ε = 1 3. So we have \|f (a) − φ(a)\| < ε = and \|f (b) − φ(b)\| < ε = 1 3 1 3 , . (19) (20) Now 1 = \|f (b) − f (a)\| = \|(f (b) − φ(b)) + (φ(b) − f (a))\| = \|(f (b) − φ(b)) + (φ(a) − f (a))\|, as φ(a) = φ(b) \|f (b) − φ(b)\| + \|φ(a) − f (a)\| < 2 3 , by (19) and (20). As it is not true that 1 < 2 3, we have a contradiction and so our supposition that S does not separate points of C(X, F ) is false, which completes the proof of the proposition. 376 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM We now deﬁne the notion of an algebra over F , where F = R or F = C. It is roughly speaking a vector space with a multiplication of vectors. So an algebra is a set together with operations of addition, scalar multiplication, and multiplication. Let F be the ﬁeld R or C. An algebra A over F is 12.2.4 Deﬁnitions. a vector space over F together with a multiplication, that is a binary operation · : A × A → A such that for any x, y, z ∈ A and α, β ∈ F : (i) (x + y; (ii) x · (y + z) = x · y + x · z; and (iii) (αx) · (βy) = (αβ)(x · y). The algebra A is said to be a unital algebra over F if there exists an identity element I ∈ A such that I · x = x · I = x, for all x ∈ A. The algebra A is said to be an associative algebra over F if x·(y ·z) = (x·y)·z, for all x, y, z ∈ A. The algebra A is said to commutative if x · y = y · x, for all x, y ∈ A. 12.2.5 Example. real number entries is an algebra over R, where the multiplication · multiplication. This algebra is unital and associative but not commutative (Exercises For each n ∈ N, the set Mn of all n × n matrices with is matrix 12.2 #2). 12.2.6 Example. as follows: for f, g ∈ C(X, R) and α ∈ R, f + g, αf , and f · g are given by C(X, R) is an algebra over the vector space R if we deﬁne · (f + g)(x) = f (x) + g(x), for all x ∈ R (αf )(x) = α(f (x), for all x ∈ R (f · g)(x) = f (x)g(x), for all x ∈ R. Indeed C(X, R) is a commutative associative unital algebra, where the identity I is the function f (x) = 1, for all x ∈ R. 12.2. THE STONE-WEIERSTRASS THEOREM 377 12.2.7 Deﬁnition. If S is itself an algebra with the operations of addition of vectors, scalar multiplication, and the binary operation · of A, then S is called a subalgebra of A. Let S be a subset of an algebra A. 12.2.8 Example. (i) The set P of all polynomials on [0, 1] is a unital subalgebra of the algebra (C[0, 1], R); (ii) The set of all Bernstein polynomials on [0, 1] is a unital subalgebra of the algebra (C[0, 1], R); (iii) Let A be a subalgebra of (C[0, 1], R) containing the identity function fI and the constant function f1, where fI is given by fI (x) = x, for all x ∈ [0, 1], and the constant function f1 is given by f1(x) = 1, for all x ∈ [0, 1]. Then A contains P [Exercise]. (iv) It follows immediately from (iii) above and Proposition 12.1.9 (i) that the algebra A in (iii) above is dense in (C[0, 1], R). Let A be an associative algebra over F , where F is R 12.2.9 Deﬁnitions. or C. Let A also be a normed vector space over F with norm \|\| \|\|. Then A is a normed algebra if for all x, y ∈ A, \|\|x · y\|\| \|\|x\|\| \|\|y\|\|. If the norm \|\| \|\| is also complete, that is A is a Banach space, then the normed algebra is called a Banach algebra. 12.2.10 Remark. multiplication · : A × A → A is a continuous mapping. it is readily seen that if A is a normed algebra, then the For any n ∈ N, and F equal to R or C, F n with the 12.2.11 Example. norm \|\|x\|\| = \|x\|, for all x ∈ F , is a Banach algebra if multiplication is given by (x1, x2, . . . , xn)·(y1, y2, . . . , yn) = (x1y1, x2y2, . . . , xnyn), for (x1, x2, . . . , xn), (y1, y2, . . . , yn) ∈ F n. (Exercise). 378 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM Let X be a compact Hausdorﬀ space and F equal to 12.2.12 Proposition. either C or R. Then C(X, F ) is a unital Banach algebra if for each f ∈ C(X, F ), \|\|f \|\| = sup x∈X \|f (x)\|. Proof. Exercise. For completeness here, we repeat the deﬁnition of a partial order in a set which appeared in Deﬁnitions 10.2.1 (See Davey and Priestley [97].) A partial order on a set X is a binary relation, 12.2.13 Deﬁnition. denoted by , which has the properties: (i) x x, for all x ∈ X (reﬂexive) (ii) if x y and y x, then x = y, for x, y ∈ X (antisymmetric), and (iii) if x y and y z, then x z, for x, y, z ∈ X (transitive) The set X equipped with the partial order is called a partially ordered set and denoted by (X, ). If x y and x = y, then we write x < y. 12.2. THE STONE-WEIERSTRASS THEOREM 379 Let X be a non-empty partially ordered set and let 12.2.14 Deﬁnitions. S ⊆ X. An element u ∈ X is said to be an upper bound of S if s u, for every s ∈ S. The element l ∈ X is said to be a lower bound of S if l s, every s ∈ S. for the element x ∈ X is an upper bound of S and is If such that x u for each upper bound u of S, then x is said to be the least upper bound of S. the element x ∈ X is a lower bound of S and is such that then x is said to be the greatest If l x for each lower bound l of S, lower bound of S. Let x, y ∈ X and put S = {x, y}. then it is denoted by x ∨ y. If the least upper bound of S exists, Let x, y ∈ X and put S = {x, y}. exists, then it is denoted by x ∧ y. If the greatest lower bound of S If x ∨ y and x ∧ y exist for all x, y ∈ X, then X is said to be a lattice. Let L be a lattice with partial order . If S is a subset of L, then S is said to be a sublattice of L if S with the partial order is also a lattice. Let X be any compact Hausdorﬀ space and f, g ∈ C(X, R). 12.2.15 Example. Then C(X, R) is a partially ordered set if f g is deﬁned to mean f (x) g(x), for each x ∈ X. Indeed, noting that (f ∨ g)(x) = max{f (x), g(x)} (f ∧ g)(x) = min{f (x), g(x)}, we see that C(X, R) is a lattice. 380 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12.2.16 Remark. We now begin our proof of the Stone-Weierstrass Theorem in earnest. The proof follows that in Simmons [362]. The ﬁrst step is to prove a version of the Stone-Weierstrass Theorem for certain sublattices of C(X, R). Then we use the Weierstrass Approximation Theorem 12.1.5 to prove that any closed subalgebra of C(X, R) is a sublattice of C(X, R). Then it is easy to prove the Stone-Weierstrass Theorem for C(X, R). Finally we prove the Stone-Weierstrass Theorem for C(X, C). Let X be a compact Hausdorﬀ space with more than 12.2.17 Proposition. one point and let L be a closed sublattice of C(X, R). If L has the property that for any x, y ∈ X with x = y and any a, b ∈ R, there exists φ ∈ L such that φ(x) = a and φ(y) = b, then L = C(X, R). Proof. Let f ∈ C(X, R). We need to show that f ∈ L. Let ε > 0 be given. Since L is closed in C(X, R), it suﬃces to ﬁnd a function g ∈ L such that \|f (x) − g(x)\| < ε, for all x ∈ X, as this implies that \|\|f − g\|\| = sup{\|f (x) − g(x)\| : x ∈ X} < ε and so L is dense (and closed) in C(X, R). Fix a point x ∈ X and let y ∈ X, y = x. By our assumption on L, there exists a function φy ∈ L such that φy(x) = f (x) and φy(y) = f (y). Let Oy be the open set given by Oy = {t : t ∈ X, φy(t) < f (t) + ε}. Clearly x, y ∈ Oy. So {Oy : y ∈ X} is an open covering of the compact space X. So there is a ﬁnite subcover O1, O2, . . . , On of X. If the corresponding functions in L are denoted by φ1, φ2, . . . , φn, the function Φx = φ1 ∧ φ2 ∧ · · · ∧ φn ∈ L and Φx(x) = x and Φx(t) < f (t) + ε, for all t ∈ X. For each x ∈ X, deﬁne the open set Ux = {t : t ∈ X, Φx(t) > f (t) − ε}. Since x ∈ Ux, the sets Ux, x ∈ X are an open covering of the compact space X. So there is a ﬁnite subcover U1, U2, . . . , Um of X. We denote the corresponding functions in L by Φ1, Φ2, . . . , Φm. Deﬁne a function g ∈ L by g = Φ1 ∨ Φ2 · · · ∨ Φm. 12.2. THE STONE-WEIERSTRASS THEOREM 381 It is clear that f (t) − ε < g(t) < f (t) + ε, for all t ∈ X, which completes our proof. 12.2.18 Lemma. and \|f \| deﬁned by \|f \|(x) = \|f (x)\|, for every x ∈ X. Let X be any compact Hausdorﬀ space, f ∈ C(X, R) (i) \|f \| ∈ C(X, R); (ii) f ∧ g = 1 (iii) f ∨ g = 1 2(f + g − \|f − g\|), for all f, g ∈ C(X, R); 2(f + g + \|f − g\|), for all f, g ∈ C(X, R); (iv) let A be any vector subspace of C(X, R) with the property that f ∈ A =⇒ \|f \| ∈ A. Then A is a sublattice of C(X, R). Proof. Exercise 12.2.19 Proposition. closed subalgebra of C(X, R) is a closed sublattice of C(X, R). Let X be a compact Hausdorﬀ space. Then every Let A be a closed subalgebra of C(X, R). If f, g ∈ C(X, R). By Lemma Proof. 7.2.18 (iv) it suﬃces to show that f ∈ A =⇒ \|f \| ∈ A. Let ε > 0 be given. For any f ∈ C(X, R), deﬁne the the closed interval [a, b] ⊂ R, where a = −\|\|f \|\| and b = \|\|f \|\|. The function φt : [a, b] → R given from [a, b] into R. So by the by φt = \|t\|, for t ∈ [a, b] is a continuous function. Weierstrass Theorem 12.1.5, there exists a polynomial p such that \| \|t\| − p(t) \| < ε 2 ,
for every t ∈ [a, b]. Deﬁne the polynomial p by p(t) = p(t) − p(0), for all t ∈ [a, b], and note that \|p(0)\| < ε \| \|t\| − p(t) \| = \| \|t\| − p(t) + p(0) \| \| \|t\| − p(t) \| + \|p(0)\| < ε, for every t ∈ [a, b]. 2 . So we have p(0) = 0 and Since A is an algebra, the function p(f ) ∈ C(X, R), given by (p(f ))(x) = p(f (x)) for all x ∈ X, is in A. Since f (x) ∈ [a, b], for all x ∈ X, the previous paragraph implies that \| \|f (x)\|−p(f (x)) \| < ε, for all x ∈ X. So \|\| \|f \|−p(f ) \|\| < ε. As p(f ) ∈ A and A is a closed set, this implies that \|f \| ∈ A, which completes the proof. 382 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM [The (Real) Stone-Weierstrass Theorem] Let X 12.2.20 Theorem. be a compact Hausdorﬀ space and A a closed subalgebra of C(X, R) which contains a non-zero constant function. Then A = C(X, R) if and only if A separates points of X. Firstly, from Proposition 12.2.3, if A = C(X, R) then A must separate Proof. points of X. We now consider the case that A separates points of X. If X has only one point, then each f ∈ C(X, R) is a constant function and as A contains a non-zero constant function and is an algebra, it contains all constant functions and so equals C(X, R). So consider the case that X has more than one point. By Proposition 12.2.17 and Proposition 12.2.19 it suﬃces to show that if x, y ∈ X with x = y and a, b ∈ R, there exists f ∈ A such that f (x) = a and f (y) = b. As A separates points of X, there exis a g ∈ A, such that g(x) = g(y). So we deﬁne f : X → R by f (z) = a g(z) − g(y) g(x) − g(y) + b g(z) − g(x) g(y) − g(x) , for z ∈ X. Then f has the required properties, which completes the proof. If f : X → C is any function from a set X into C, then we 12.2.21 Remark. can write f (x) = R(f )(x) + i I(f )(x), where R(f )(x) ∈ R and I(f )(x) ∈ R. So f = R(f ) + i I(f ). The conjugate function, f is deﬁned to be R(f ) − i I(f ). Note that R(f ) = f + f 2 and I(f ) = f − f 2i . 12.2. THE STONE-WEIERSTRASS THEOREM 383 [The (Complex) Stone-Weierstrass Theorem] 12.2.22 Theorem. Let X be a compact Hausdorﬀ space and A a closed subalgebra of C(X, C) which contains a non-zero constant function. Then A = C(X, C) if and only if f ∈ A implies the conjugate function f ∈ A and A separates points of X. From Proposition 12.2.3, if A = C(X, C), A must separate points of X Proof. and obviously f ∈ A = C(X, C) implies the conjugate function f ∈ A. So we consider the converse statement. Deﬁne B to be the real-valued functions in A. Clearly B is a closed subalgebra of C(X, R). We claim that it suﬃces to prove that B = C(X, R). This is so, since if f ∈ C(X, C), then R(f ) and I(f ) are in C(X, R) and so in B = C(X, R), and so A would be C(X, C). We shall use the Real Stone-Weierstrass Theorem 12.2.20 to prove that B = C(X, R). Let f ∈ C(X, C) seprate points of X. Then either R(f ) or I(f ) (or both) separate points of X; so B separates points of X. Now A contains a non-constant function g. As A is an algebra, the conjugate function g is also in A. So the non-constant real-valued function gg = \|g\|2 ∈ B. So by the Real Stone-Weierstrass Theorem 12.2.20, B = C(X, R), which completes the proof. 7.2.23 Remark. The Weierstrass Approximation Theorem 12.1.5 is of course a special case of The (Complex) Stone-Weierstrass Theorem 12.2.22. It is the case X = [a, b] ⊂ R and A is the set of all polynomials. 384 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12.2.24 Remark. We mentioned at the beginning of §12.1 that in 1885 Karl Weierstrass was aged 70, when he proved the The Weierstrass Approximation Theorem 12.1.5. In contrast, in 1899 the Hungarian Jewish mathematician Lipót Fejér (1880–1959), aged 19, proved the trignometric polynomial version of this theorem from which The Weierstrass Approximation Theorem 12.1.5 can be derived. Lipót Fejér was born Lipót Weiss and changed his name in high school as he expected less antisemitism. Research students that he Fejér supervized included: Marcel Riesz, George Pólya, Gábor Szegő, John von Neumann, Pál Turán, and Paul Erdős - an incredible heritage. Riesz Pólya Szegő von Neumann Turán Erdős Neumannsmall.jpg Let a0, an, bn ∈ R, n = 1, 2 with either 12.2.25 Deﬁnition. aN = 0 or bN = 0. Then the function f : R → R given by N f (x) = a0 + an cos(nx) + bn sin(nx), for x ∈ R is said to be a real trignometric polynomial of degree N . n=1 12.2. THE STONE-WEIERSTRASS THEOREM 385 Let a0, an, bn ∈ C, n = 1, 2 with either 12.2.26 Deﬁnition. aN = 0 or bN = 0. Then the function f : R → C given by N f (x) = a0 + an cos(nx) + i bn sin(nx), for x ∈ R is is said to be a complex trignometric polynomial of degree N . n=1 12.2.27 Remark. We observe that if f is a real trignometric polynomial or a complex trignometric polynomial then f (x) = f (x + 2π) = f (x + 2kπ), for all k ∈ N and x ∈ R. (21) We shall prove that every function f : R → R satisfying (21) can be approximated by real trignometric polynomials and every function f : R → C satisfying (21) can be approximated by complex trignometric polynomials. Let T denote the circle of diameter one centred at 0 in the 12.2.29 Remark. euclidean space R2 with the subspace topology from R2. Of course T is a compact Hausdorﬀ space. Further, any function f : R → R or f : R → C which satisﬁes (21) can be thought of as a function from T to R or from T to C. Further, f is a continuous function if and only if the corresponding function from T to R or C is continuous. The algebra of all real trignometric polynomials is dense 12.2.30 Corollary. in C(T, R) and the algebra of all complex trignometric polynomials is dense in C(T, C). Proof. These results follow from The Real Stone-Weierstrass Theorem 12.2.20 and The Complex Stone-Weierstrass Theorem 12.2.22. [Exercise] 386 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 12.2.31 Remark. Lord Kelvin4 (see Kelvin [234]), was interested in correcting the magnetic compass in ships, which usually had a lot of iron and steel, in order to get a true north reading and was also interested in predicting tides. Both of these problems were able to be addressed using trignometric polynomial approximation. For a discussion see Sury Kelvin [387]. We conclude this section with an interesting remark, unrelated to the Stone- Weierstrass Theorem, on the spaces C(X, R). 12.2.31 Remark. While C([0, 1], R) is just one separable Banach space, the Indeed it is a following result says it is much richer than you may have thought. universal separable metric space, as described below. Banach Mazur Theorem. Every separable Banach space is isometrically embeddable as a metric space in C[(0, 1), R). From the Banach-Alaoglu Theorem, Exercises 10.3 #33 (vii) and the Hahn-Banach Theorem5 the following beautiful generalization can be deduced. If B is any Banach space, then there exists a compact Hausdorﬀ space X such that B is isometrically embeddable as a metric space in C(X, R). (See Maddox [271], Theorem 27.) As a generalization of the concept of separability of a topological space, we introduce the following notion. The density character of a topological space X is the least cardinal number of a dense subspace of X. In 1969 Kleiber and Pervin [239] proved the next theorem: The topological space C([0, 1]ℵ, R) with the uniform topology is a universal that is, C([0, 1]ℵ, R) has density metric space of density character ℵ; character ℵ and every metric space of density character ℵ is isometrically embedded as a subspace of C([0, 1]ℵ, R). 4William Thompson (1824-1907), born in Belfast, Ireland, was the son of the Professor of Mathematics at Glasgow University. He attended university classes from the age of 10. He graduated from Camridge University and at the age of 22 returned to Glasgow University to become Professor of Natural Sciences, a position he held for over half a century. He became a Lord in 1892, and took the name Kelvin. 5https://tinyurl.com/zj9byaj 12.2. THE STONE-WEIERSTRASS THEOREM 387 Exercises 12.2 1. Verify that the subset S of C([0, 1], R) consisting of all polynomials in C[0, 1] with rational number coeﬃcients is not a subalgebra of C[0, 1]. 2. Prove that the set Mn of all n×n matrices with real number entries is an algebra over R in Example 12.2.3 is a non-commutative associative unital algebra over R. 3. Let X be a compact Hausdorﬀ space, A a subalgebra of C(X, F ), where F is C or R, and B the closure in C(X, F ) of A. Prove that (i) B is a subalgebra of C(X, F ); (ii) if A contains the conjugate of each of its functions, then so does B. 4. Verify Example 12.2.6 (iii). 5. Verify Example 7.2.9 is correct. 6. Prove Proposition 12.2.10. 7. Verify Lemma 7.2.16. 8. Let X be a locally compact Hausdorﬀ space and f a continuous function from X into F , where F is C or R. Then f is said to vanish at inﬁnity if for each ε > 0, there exists a compact subset K of X such that \|f (x)\| < ε, for all x ∈ X \ K. Let C0(X, F ) be the Banach algebra of all continuous functions which vanish at inﬁnity. Prove that C0(X, F ) with the sup norm is a Banach algebra and is unital Banach algebra if and only if X is compact. 388 CHAPTER 12. THE STONE-WEIERSTRASS THEOREM 9. Let A be a Banach algebra over C. Then A is said to be a C∗-algebra6 7 if there is a map ∗ : A → A which satisﬁes (i)–(v). (i) x∗∗ = (x∗)∗ = x, for every x ∈ A; that is ∗ is an involution; (ii) (x + y)∗ = x∗ + y∗, for every x, y ∈ A; (iii) (x · y)∗ = y∗ · x∗, for every x, y ∈ A; (iv) (αx)∗ = αx∗, for every x ∈ A and α ∈ C, where α denotes the complex conjugate of the complex number α; (v) \|\|x∗ · x\|\| = \|\|x\|\|2, for every x ∈ A. For any x ∈ A, x∗ is called the adjoint of x. (a) Using the fact that in any Banach algebra A, \|\|x · y\|\| \|\|x\|\| \|\|y\|\|, for all x, y ∈ A, verify that \|\|x\|\| = \|\|x∗\|\|. (b) Prove that if X is a compact Hausdorﬀ space, then the unital commutative associative Banach algebra C(X, C) is a C∗-algebra if for each f ∈ C(X, C) we deﬁne f ∗ : X → C by f ∗(x) = f (x), for every x ∈ X, where f (x) denotes the complex conjugate of the complex number f (x). Let A and B be C∗-algebras. Then A is said to be is
omorphic as a C∗algebra to B if there is a surjective one-to-one mapping φ : A → B such that φ(x + y) = φ(x) + φ(y), φ(αx) = αφ(x), φ(x · y) = φ(x) · φ(y), and φ(x∗) = (φ(x))∗, for all x, y ∈ A and α ∈ C. The Gelfand-Naimark Representation Theorem, proved using the Stone-Weierstrass Theorem, says every commutative unital C∗-algebra A is isomorphic as a C∗-algebra to C(X, C), for some compact Hausdorﬀ space X. More generally, every commutative C∗-algebra A is isomorphic as a C∗algebra to C0(X, C), for some locally compact Hausdorﬀ space X. (See Lang [252], Chapter 16, Theorem 3.3.) It is also worth mentioning that C(X, C) is isomorphic as a C∗-algebra to C(Y, C), for compact Hausdorﬀ spaces X and Y , if and only if X is homeomorphic to Y . (See Lin [261], Theorem 1.3.9.) 6The concept of a C∗-algebra has its roots in the work on quantum mechanics of Werner Karl Heisenberg (1901–1976), Erwin Rudolf Josef Alexander Schrödinger (1887-1961), and John von Neumann (1903–1957) in the 1920s. (See Landsman [251].) 7Gelfand and Naimark [151] 12.3. CREDIT FOR IMAGES 12.3 Credit for Images 1. Jacob Bernoulli. Public Domain. 2. Sergei Natanovich Bernstein. Public Domain. 389 3. Pierre Bezier portrait. Creative Commons. https://commons.wikimedia.org/wiki/ File:Pierre-bezier-portrait.jpg 4. Ulisse Dini. Public Domain. 5. Paul Erdős. Photo taken by author November 23, 1979 at the University of Calgary. 6. Lipót Fejér. Public Domain. 7. Lord Kelvin (William Thompson). Public Domain 8. George Pólya. Public Domain. 9. Marcel Riesz. Public Domain. 10. Marshall Stone. Public Domain. 11. Gábor Szegő. Public Domain. 12. Pál Turán. Public Domain. 13. John von Neumann. Public Domain. 14. Rudolf Výborný. Privately owned photo. 15. Karl Weierstrass. Public Domain. Appendix 1: Inﬁnite Sets §A1.0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 §A1.1 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 §A1.2 Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404 §A1.3 Cardinal Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 §A1.4 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 §A1.5 Credits for Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 390 A1.0 Introduction 391 “The Inﬁnite! No other question has ever moved so profoundly the spirit of man.” - David Hilbert (1921) Once upon a time in a far-oﬀ land there were two hotels, the Hotel Finite (an ordinary hotel with a ﬁnite number of rooms) and Hilbert’s Hotel Inﬁnite (an extraordinary hotel with an inﬁnite number of rooms numbered 1, 2, . . . n, . . . ). One day a visitor arrived in town seeking a room. She went ﬁrst to the Hotel Finite and was informed that all rooms were occupied and so she could not be accommodated, but she was told that the other hotel, Hilbert’s Hotel Inﬁnite, can always ﬁnd an extra room. So she went to Hilbert’s Hotel Inﬁnite and was told that there too all rooms were occupied. However, the desk clerk said at this hotel an extra guest can always be accommodated without evicting anyone. He moved the guest from room 1 to room 2, the guest from room 2 to room 3, and so on. Room 1 then became vacant! From this cute example we see that there is an intrinsic diﬀerence between inﬁnite sets and ﬁnite sets. The aim of this Appendix is to provide a gentle but very brief introduction to the theory of Inﬁnite Sets8. This is a fascinating topic which, if you have not studied it before, will contain several surprises. We shall learn that “inﬁnite sets were not created equal" - some are bigger than others. At ﬁrst pass it is not at all clear what this statement could possibly mean. We will need to deﬁne the term “bigger". Indeed we will need to deﬁne what we mean by “two sets are the same size". 8There is available for free download a rather nice and gentle book on set theory. It is by Raymond L. Wilder and is called Introduction to the Foundations of Mathematics. It is available from http://archive.org/details/IntroductionToTheFoundationsOfMathematics 392 APPENDIX 1: INFINITE SETS There are three videos which you should watch as they provide supplementary material to this Appendix. These videos are called “Topology Without Tears – Video 2a, 2b, and 2c – Inﬁnite Set Theory”. Part (a) is on YouTube at http://youtu.be/9h83ZJeiecg and on the Chinese Youku site at http://tinyurl.com/ m4dlzhh. Part (b) is on YouTube at http://youtu.be/QPSRB4Fhzko and on the Chinese Youku site at http://tinyurl.com/ kf9lp8e. Part (c) is on YouTube at http://youtu.be/YvqUnjjQ3TQ and on the Chinese Youku site at http://tinyurl.com/ mhlqe93. These videos include a discussion of the Zermelo-Fraenkel (ZF) axioms of Set Theory and a short proof showing that the Russell Paradox does not occur within ZF set theory. A1.1 Countable Sets 393 Let A and B be sets. Then A is said to be equipotent A1.1.1 Deﬁnitions. to B, denoted by A ∼ B, if there exists a function f : A → B which is both oneto-one and onto (that is, f is a bijection or a one-to-one correspondence). A1.1.2 Proposition. Let A, B, and C be sets. (i) Then A ∼ A. (ii) If A ∼ B then B ∼ A. (iii) If A ∼ B and B ∼ C then A ∼ C. Outline Proof. (i) The identity function f on A, given by f (x) = x, for all x ∈ A, is a one-to-one correspondence between A and itself. (ii) If f is a bijection of A onto B then it has an inverse function g from B to A and g is also a one-to-one correspondence. (iii) If f : A → B is a one-to-one correspondence and g : B → C is a one-toone correspondence, then their composition gf : A → C is also a one-to-one correspondence. Proposition A1.1.2 says that the relation “∼” is reﬂexive (i), symmetric (ii), and transitive (iii); that is, “∼” is an equivalence relation. A1.1.3 Proposition. {1, 2, . . . , m} are equipotent if and only if n = m. Let n, m ∈ N. Then the sets {1, 2, . . . , n} and Proof. Exercise. Now we explicitly deﬁne the terms “ﬁnite set” and “inﬁnite set”. 394 APPENDIX 1: INFINITE SETS A1.1.4 Deﬁnitions. Let S be a set. (i) Then S is said to be ﬁnite if it is the empty set, Ø, or it is equipotent to {1, 2, . . . , n}, for some n ∈ N. (ii) If S is not ﬁnite, then it is said to be inﬁnite. (iii) If S ∼ {1, 2, . . . , n} then S is said to have cardinality n, which is denoted by card S = n. (iv) If S = Ø then the cardinality is said to be 0, which is denoted by card Ø = 0. The next step is to deﬁne the “smallest” kind of inﬁnite set. Such sets will be called countably inﬁnite. At this stage we do not know that there is any “bigger” kind of inﬁnite set – indeed we do not even know what “bigger” would mean in this context. A1.1.5 Deﬁnitions. Let S be a set. (i) The set S is said to be countably inﬁnite (or denumerable) if it is equipotent to N. (ii) The set S is said to be countable if it is ﬁnite or countably inﬁnite. (iii) If S is countably inﬁnite then it is said to have cardinality ℵ0, denoted by card S = ℵ0. (iv) A set S is said to be uncountable if it is not countable. If the set S is countably inﬁnite, then S = {s1, s2, . . . , sn, . . . } A1.1.6 Remark. where f : N → S is a one-to-one correspondence and sn = f (n), for all n ∈ N. So we can list the elements of S. Of course if S is ﬁnite and non-empty, we can also list its elements by S = {s1, s2, . . . , sn}. So we can list the elements of any countable set. Conversely, if the elements of S can be listed, then S is countable as the listing deﬁnes a one-to-one correspondence with N or {1, 2, . . . , n}. 395 A1.1.7 Example. The set S of all even positive integers is countably inﬁnite. Proof. The function f : N → S given by f (n) = 2 n, for all n ∈ N, is a one-to-one correspondence. Example A1.1.7 is worthy of a little contemplation. We think of two sets being in one-to-one correspondence if they are “the same size”. But here we have the set N in one-to-one correspondence with one of its proper subsets. This does not happen with ﬁnite sets. Indeed ﬁnite sets can be characterized as those sets which are not equipotent to any of their proper subsets. A1.1.8 Example. The set Z of all integers is countably inﬁnite. Proof. The function f : N → Z given by f (n) =    m, −m, 0, if n = 2m, m 1 if n = 2m + 1, m 1 if n = 1. is a one-to-one correspondence. A1.1.9 Example. The set S of all positive integers which are perfect squares is countably inﬁnite. Proof. The function f : N → S given by f (n) = n2 is a one-to-one correspondence. Example A1.1.9 was proved about 1600 by the Italian astronomer, engineer, and physicist Galileo Galilei (1564–1642). It troubled him and suggested to him that the inﬁnite is not man’s domain. Gailieo has been desctribed as the "father of modern physics", the "father of the scientiﬁc method", and the "father of modern science”. Galileo A1.1.10 Proposition. If a set S is equipotent to a countable set then it is countable. Proof. Exercise. 396 APPENDIX 1: INFINITE SETS A1.1.11 Proposition. If S is a countable set and T ⊂ S then T is countable. Proof. if S is ﬁnite, an inﬁnite one if S is countably inﬁnite). Since S is countable we can write it as a list S = {s1, s2, . . .} (a ﬁnite list Let t1 be the ﬁrst si in T (if T = Ø). Let t2 be the second si in T (if T = {t1}). Let t3 be the third si in T (if T = {t1, t2}), . . . . This process comes to an end only if T = {t1, t2, . . . , tn} for some n, in which case T is ﬁnite. If the process does not come to an end we obtain a list {t1, t2, . . . , tn, . . .} of members of T . This list contains every member of T , because if si ∈ T then we reach si no later than the ith step in the process; so si occurs in the list. Hence T is countably inﬁnite. So T is either ﬁnite or countably inﬁnite. As
an immediate consequence of Proposition A1.1.11 and Example A1.1.8 we have the following result. A1.1.12 Corollary. Every subset of Z is countable. A1.1.13 Lemma. If S1, S2, . . . , Sn, . . . is a countably inﬁnite family of countably inﬁnite sets such that Si ∩ Sj = Ø for i = j, then inﬁnite set. ∞ i=1 Si is a countably As each Si is a countably inﬁnite set, Si = {si1, si2, . . . , sin, . . .}. Now Proof. put the sij in a square array and list them by zigzagging up and down the short diagonals. s11 → s12 s22 s21 ↓ s31 ... s23 s13 → s14 · · · · · · · · · . . . s33 ... s32 ... i=1 Si are listed, and the list is inﬁnite because This shows that all members of ∞ each Si is inﬁnite. So ∞ i=1 Si is countably inﬁnite. In Lemma A1.1.13 we assumed that the sets Si were pairwise disjoint. If they are not pairwise disjoint the proof is easily modiﬁed by deleting repeated elements to obtain: 397 A1.1.14 Lemma. If S1, S2, . . . , Sn, . . . is a countably inﬁnite family of countably inﬁnite sets, then Si is a countably inﬁnite set. ∞ i=1 A1.1.15 Proposition. The union of any countable family of countable sets is countable. Proof. Exercise. A1.1.16 Proposition. product set S × T = {s, t : s ∈ S, t ∈ T } is a countably inﬁnite set. If S and T are countably inﬁnite sets then the Proof. Let S = {s1, s2, . . . , sn, . . . } and T = {t1, t2, . . . , tn, . . . }. Then ∞ S × T = {si, t1, si, t2, . . . , si, tn, . . . }. i=1 So S × T is a countably inﬁnite union of countably inﬁnite sets and is therefore countably inﬁnite. A1.1.17 Corollary. Every ﬁnite product of countable sets is countable. We are now ready for a signiﬁcant application of our observations on countable sets. A1.1.18 Lemma. The set, Q>0, of all positive rational numbers is countably inﬁnite. Let Si be the set of all positive rational numbers (see Niven [317]) with i , . . . , n Si. As each Proof. i , 2 denominator i, for i ∈ N. Then Si = Si is countably inﬁnite, Proposition A1.1.15 yields that Q>0 is countably inﬁnite. and Q>0 = ∞ i=1 i , . . . 1 398 APPENDIX 1: INFINITE SETS We are now ready to prove that the set, Q, of all rational numbers is countably inﬁnite; that is, there exists a one-to-one correspondence between the set Q and the (seemingly) very much smaller set, N, of all positive integers. A1.1.19 Theorem. The set Q of all rational numbers is countably inﬁnite. Clearly the set Q<0 of all negative rational numbers is equipotent to the Proof. set, Q>0, of all positive rational numbers and so using Proposition A1.1.10 and Lemma A1.1.18 we obtain that Q<0 is countably inﬁnite. Finally observe that Q is the union of the three sets Q>0, Q<0 and {0} and so it too is countably inﬁnite by Proposition A1.1.15. A1.1.20 Corollary. Every set of rational numbers is countable. Proof. This is a consequence of Theorem A1.1.19 and Proposition A1.1.11. A real number x is said to be an algebraic number A1.1.21 Deﬁnitions. if there is a natural number n and integers a0, a1, . . . , an with a0 = 0 such that a0xn + a1xn−1 + · · · + an−1x + an = 0. A real number which is not an algebraic number is said to be a transcendental number. A1.1.22 Example. Every rational number is an algebraic number. If x = p Proof. number with n = 1, a0 = q, and an = −p. q , for p, q ∈ Z and q = 0, then qx − p = 0; that is, x is an algebraic A1.1.23 Example. The number √ 2 is an algebraic number which is not a rational number. Proof. While x = √ 2 is irrational, it satisﬁes x2 − 2 = 0 and so is algebraic. 399 It is also easily veriﬁed that 4√ √ 5 − A1.1.24 Remark. since it satisﬁes x8 − 12x6 + 44x4 − 288x2 + 16 = 0. which can be constructed from the set of integers using only a ﬁnite number of 3 is an algebraic number Indeed any real number the operations of addition, subtraction, multiplication, division and the extraction of square roots, cube roots, . . . , is algebraic. A1.1.25 Remark. Remark A1.1.24 shows that “most” numbers we think of are algebraic numbers. To show that a given number is transcendental can be extremely diﬃcult. The ﬁrst such demonstration was in 1844 when Liouville proved the transcendence of the number ∞ n=1 1 10n! = 0.11000100000000000000000100 . . . It was Charles Hermite who, in 1873, showed that e is transcendental. In 1882 Lindemann proved that the number π is transcendental thereby answering in the negative the 2,000 year old question about squaring the circle. (The question is: given a circle of radius 1, is it possible, using only a straight edge and compass, to construct a square with the same area? A full exposition of this problem and proofs that e and π are transcendental are to be found in the book, Jones, Morris, and Pearson [221].) We now proceed to prove that the set A of all algebraic numbers is also countably inﬁnite. This is a more powerful result than Theorem A1.1.19 which is in fact a corollary of this result. 400 APPENDIX 1: INFINITE SETS A1.1.26 Theorem. The set A of all algebraic numbers is countably inﬁnite. Consider the polynomial f (x) = a0xn + a1xn−1 + · · · + an−1x + an , where Proof. a0 = 0 and each ai ∈ Z and deﬁne its height to be k = n + \|a0\| + \|a1\| + · · · + \|an\|. For each positive integer k, let Ak be the set of all roots of all such polynomials of height k. Clearly A = ∞ k=1 Ak. Therefore, to show that A is countably inﬁnite, it suﬃces by Proposition A1.1.15 to show that each Ak is ﬁnite. If f is a polynomial of degree n, then clearly n k and \|ai\| k, for i = 0, 1, 2, . . . , n. So the set of all polynomials of height k is certainly ﬁnite. Further, a polynomial of degree n has at most n roots. Consequently each polynomial of height k has no more than k roots. Hence the set Ak is ﬁnite, as required. A1.1.27 Corollary. Every set of algebraic numbers is countable. Note that Corollary A1.1.27 has as a special case, Corollary A1.1.20. So far we have not produced any example of an uncountable set. Before doing so we observe that certain mappings will not take us out of the collection of countable sets. A1.1.28 Proposition. Let X and Y be sets and f a map of X into Y . (i) If X is countable and f is surjective (that is, an onto mapping), then Y is countable. (ii) If Y is countable and f is injective (that is, a one-to-one mapping), then X is countable. Proof. Exercise. 401 A1.1.29 Proposition. subsets of S is also countable. Let S be a countable set. Then the set of all ﬁnite Proof. Exercise. A1.1.30 Deﬁnition. to be the power set of S and is denoted by P(S). Let S be any set. The set of all subsets of S is said A1.1.31 Theorem. (Georg Cantor) For every set S, the power set, P(S), is not equipotent to S; that is, P(S) ∼ S. Proof. We have to prove that there is no one-to-one correspondence between S and P(S). We shall prove more: that there is not even any surjective function mapping S onto P(S). Suppose that there exists a function f : S → P(S) which is onto. For each x ∈ S, f (x) ∈ P(S), which is the same as saying that f (x) ⊆ S. Let T = {x : x ∈ S and x ∈ f (x)}. Then T ⊆ S; that is, T ∈ P(S). So T = f (y) for some y ∈ S, since f maps S onto P(S). Now y ∈ T or y ∈ T . Case 1. y ∈ T ⇒ y ∈ f (y) (by the deﬁnition of T ) ⇒ y ∈ T (since f (y) = T ). So Case 1 is impossible. Case 2. y ∈ T ⇒ y ∈ f (y) (by the deﬁnition of T) ⇒ y ∈ T (since f (y) = T ). So Case 2 is impossible. As both cases are impossible, we have a contradiction. So our supposition is false and there does not exist any function mapping S onto P(S). Thus P(S) is not equipotent to S. 402 APPENDIX 1: INFINITE SETS A1.1.32 Lemma. power set, P(S). If S is any set, then S is equipotent to a subset of its Deﬁne the mapping f : S → P(S) by f (x) = {x}, for each x ∈ S. Clearly Proof. f is a one-to-one correspondence between the sets S and f (S). So S is equipotent to the subset f (S) of P(S). A1.1.33 Proposition. If S is any inﬁnite set, then P(S) is an uncountable set. Proof. not equipotent to S. As S is inﬁnite, the set P(S) is inﬁnite. By Theorem A1.1.31, P(S) is Suppose P(S) is countably inﬁnite. Then by Proposition A1.1.11, Lemma A1.1.32 and Proposition A1.1.10, S is countably inﬁnite. So S and P(S) are equipotent, which is a contradiction. Hence P(S) is uncountable. Proposition A1.1.33 demonstrates the existence of uncountable sets. However the sceptic may feel that the example is contrived. So we conclude this section by observing that important and familiar sets are uncountable. A1.1.34 Lemma. [1, 2) is not countable. The set of all real numbers in the half open interval 403 Proof. numbers in [1, 2) cannot be listed. (Cantor’s diagonal argument) We shall show that the set of all real Let L = {r1, r2, . . . rn . . . } be any list of real numbers each of which lies in the set [1, 2). Write down their decimal expansions: r1 =1.r11r12 . . . r1n . . . r2 =1.r21r22 . . . r2n . . . ... rm =1.rm1rm2 . . . rmn . . . ... Consider the real number a deﬁned to be 1.a1a2 . . . an . . . where, for each n ∈ N, an = 1 2 if rnn = 1 if rnn = 1. Clearly an = rnn and so a = rn, for all n ∈ N. Thus a does not appear anywhere in the list L. Thus there does not exist a listing of the set of all real numbers in [1, 2); that is, this set is uncountable. A1.1.35 Theorem. The set, R, of all real numbers is uncountable. Suppose R is countable. Then by Proposition A1.1.11 the set of all real Proof. numbers in [1, 2) is countable, which contradicts Lemma A1.1.34. Therefore R is uncountable. 404 APPENDIX 1: INFINITE SETS A1.1.36 Corollary. The set, I, of all irrational numbers is uncountable. Suppose I is countable. Then R is the union of two countable sets: I Proof. and Q. By Proposition A1.1.15, R is countable which is a contradiction. Hence I is uncountable. Using a similar proof to that in Corollary A1.1.36 we obtain the following result. A1.1.37 Corollary. The set of all transcendental numbers is uncountable. A1.2 Cardinal Numbers In the previous section we deﬁned countably inﬁnite and uncountable and suggested, without explaining what it might mean, that uncountable sets are “bigger” than countably inﬁnite se
ts. To explain what we mean by “bigger” we will need the next theorem. Our exposition is based on that in the book, Halmos [168] 405 A1.2.1 Theorem. (Cantor-Schröder-Bernstein) Let S and T be sets. If S is equipotent to a subset of T and T is equipotent to a subset of S, then S is equipotent to T . Proof. Without loss of generality we can assume S and T are disjoint. Let f : S → T and g : T → S be one-to-one maps. We are required to ﬁnd a bijection of S onto T . We say that an element s is a parent of an element f (s) and f (s) is a descendant of s. Also t is a parent of g(t) and g(t) is a descendant of t. Each s ∈ S has an inﬁnite sequence of descendants: f (s), g(f (s)), f (g(f (s))), and so on. We say that each term in such a sequence is an ancestor of all the terms that follow it in the sequence. Now let s ∈ S. things must happen: If we trace its ancestry back as far as possible one of three (i) the list of ancestors is ﬁnite, and stops at an element of S which has no ancestor; (ii) the list of ancestors is ﬁnite, and stops at an element of T which has no ancestor; (iii) the list of ancestors is inﬁnite. Let SS be the set of those elements in S which originate in S; that is, SS is the set S \ g(T ) plus all of its descendants in S. Let ST be the set of those elements which originate in T ; that is, ST is the set of descendants in S of T \ f (S). Let S∞ be the set of all elements in S with no parentless ancestors. Then S is the union of the three disjoint sets SS, ST and S∞. Similarly T is the disjoint union of the three similarly deﬁned sets: TT , TS, and T∞. Clearly the restriction of f to SS is a bijection of SS onto TS. Now let g−1 be the inverse function of the bijection g of T onto g(T ). Clearly the restriction of g−1 to ST is a bijection of ST onto TT . Finally, the restriction of f to S∞ is a bijection of S∞ onto T∞. Deﬁne h : S → T by h(s) =    f (s) g−1(s) f (s) if s ∈ SS if s ∈ ST if s ∈ S∞. Then h is a bijection of S onto T . So S is equipotent to T . 406 APPENDIX 1: INFINITE SETS Our next task is to deﬁne what we mean by “cardinal number”. A1.2.2 Deﬁnitions. A collection, ℵ, of sets is said to be a cardinal number if it satisﬁes the conditions: (i) Let S and T be sets. If S and T are in ℵ, then S ∼ T ; (ii) Let A and B be sets. If A is in ℵ and B ∼ A, then B is in ℵ. If ℵ is a cardinal number and A is a set in ℵ, then we write card A = ℵ. Deﬁnitions A1.2.2 may, at ﬁrst sight, seem strange. A cardinal number is deﬁned as a collection of sets. So let us look at a couple of special cases: If a set A has two elements we write card A = 2; the cardinal number 2 is the collection of all sets equipotent to the set {1, 2}, that is the collection of all sets with 2 elements. If a set S is countable inﬁnite, then we write card S = ℵ0; in this case the cardinal number ℵ0 is the collection of all sets equipotent to N. Let S and T be sets. Then S is equipotent to T if and only if card S = card T . A1.2.3 Deﬁnitions. c. The cardinality of N is denoted by ℵ0. The cardinality of R is denoted by c; that is, card R = The symbol c is used in Deﬁnitions A1.2.3 as we think of R as the “continuum”. We now deﬁne an ordering of the cardinal numbers. Let m and n be cardinal numbers. Then the cardinal A1.2.4 Deﬁnitions. m is said to be less than or equal to n, that is m n, if there are sets S and T such that card S = m, card T = n, and S is equipotent to a subset of T . Further, the cardinal m is said to be strictly less than n, that is m < n, if m n and m = n. As R has N as a subset, card R = c and card N = ℵ0, and R is not equipotent to N, we immediately deduce the following result. 407 A1.2.5 Proposition. ℵ0 < c. We also know that for any set S, S is equipotent to a subset of P(S), and S is not equipotent to P(S), from which we deduce the next result. A1.2.6 Theorem. For any set S, card S < card P(S). The following is a restatement of the Cantor-Schröder-Bernstein Theorem 1.2.1. A1.2.7 Theorem. then m = n. Let m and n be cardinal numbers. If m n and n m, A1.2.8 Remark. We observe that there are an inﬁnite number of inﬁnite cardinal numbers. This is clear from the fact that: (∗) ℵ0 = card N < card P(N) < card P(P(N)) < . . . The next result is an immediate consequence of Theorem A1.2.6. A1.2.9 Corollary. There is no largest cardinal number. Noting that if a ﬁnite set S has n elements, then its power set P(S) has 2n elements, it is natural to introduce the following notation. A1.2.10 Deﬁnition. P(S) is denoted by 2ℵ. If a set S has cardinality ℵ, then the cardinality of Thus we can rewrite (∗) above as: (∗∗) ℵ0 < 2ℵ0 < 22ℵ0 < 222ℵ0 < . . . . When we look at this sequence of cardinal numbers there are a number of questions which should come to mind including: 408 APPENDIX 1: INFINITE SETS (1) Is c equal to one of the cardinal numbers on this list? (2) Are there any cardinal numbers strictly between ℵ0 and 2ℵ0? These questions, especially (2), are not easily answered. Indeed they require a careful look at the axioms of set theory. It is not possible in this Appendix to discuss seriously the axioms of set theory. Nevertheless we will touch upon the above questions later in the appendix. We conclude this section by identifying the cardinalities of a few more familiar sets. A1.2.11 Lemma. (i) [0, 1] ∼ [a, b]; (ii) (0, 1) ∼ (a, b); (iii) (0, 1) ∼ (1, ∞); Let a and b be real numbers with a < b. Then (iv) (−∞, −1) ∼ (−2, −1); (v) (1, ∞) ∼ (1, 2); (vi) R ∼ (−2, 2); (vii) R ∼ (a, b). (i) is proved by observing that f (x) = a + (b − a)x deﬁnes a Outline Proof. one-to-one function of [0, 1] onto [a, b]. (ii) and (iii) are similarly proved by ﬁnding suitable functions. (iv) is proved using (iii) and (ii). (v) follows from (iv). (vi) follows from (iv) and (v) by observing that R is the union of the pairwise disjoint sets (−∞, −1), [−1, 1] and (1, ∞). (vii) follows from (vi) and (ii). . 409 A1.2.12 Proposition. is any subset of R such that (a, b) ⊆ S, then card S = c. card (a, b) = card [a, b] = c. Let a and b be real numbers with a < b. If S In particular, Proof. Using Lemma A1.2.11 observe that card R = card (a, b) card [a, b] card R. So card (a, b) = card [a, b] = card R = c. . A1.2.13 Proposition. then card (R2) = c. If R2 is the set of points in the Euclidean plane, Outline Proof. By Proposition A1.2.12, R is equipotent to the half-open interval [0, 1) and it is easily shown that it suﬃces to prove that [0, 1) × [0, 1) ∼ [0, 1). Deﬁne f : [0, 1) → [0, 1) × [0, 1) by f (x) is the point x, 0. Then f is a one-to- one mapping of [0, 1) into [0, 1) × [0, 1) and so c = card [0, 1) card [0, 1) × [0, 1). By the Cantor-Schröder-Bernstein Theorem A.2.1, it suﬃces then to ﬁnd a one-to-one function g of [0, 1) × [0, 1) into [0, 1). Deﬁne g(0.a1a2 . . . an . . . , 0.b1b2 . . . bn . . . , ) = 0.a1b1a2b2 . . . anbn . . . . Clearly g is well-deﬁned (as each real number in [0, 1) has a unique decimal representation that does not end in 99. . . 9. . . ) and is one-to-one, which completes the proof. A1.3 Cardinal Arithmetic We begin with a deﬁnition of addition of cardinal numbers. Of course, when the cardinal numbers are ﬁnite, this deﬁnition must agree with addition of ﬁnite numbers. Let α and β be any cardinal numbers and select disjoint A1.3.1 Deﬁnition. sets A and B such that card A = α and card B = β. Then the sum of the cardinal numbers α and β is denoted by α + β and is equal to card (A ∪ B). 410 APPENDIX 1: INFINITE SETS A1.3.2 Remark. Before knowing that the above deﬁnition makes sense and in particular does not depend on the choice of the sets A and B, it is necessary to verify that if A1 and B1 are disjoint sets and A and B are disjoint sets such that card A = card A1 and card B = card B1, then A ∪ B ∼ A1 ∪ B1; that is, card (A ∪ B) = card (A1 ∪ B1). This is a straightforward task and so is left as an exercise. A1.3.3 Proposition. For any cardinal numbers α, β and γ : (iii) α + (β + γ) = (α + β) + γ ; (iii) α + 0 = α ; (iv) If α β then α + γ β + γ . Proof. Exercise A1.3.4 Proposition. (i) ℵ0 + ℵ0 = ℵ0; (ii) c + ℵ0 = c; (iii) c + c = c; (iv) For any ﬁnite cardinal n, n + ℵ0 = ℵ0 and n + c = c. Proof. (i) The listing 1, −1, 2, −2, . . . , n, −n, . . . shows that the union of the two countably inﬁnite sets N and the set of negative integers is a countably inﬁnite set. (ii) Noting that [−2, −1] ∪ N ⊂ R, we see that card [−2, −1] + card N card R = c. So c = card [−2, −1] card ([−2, −1]∪N) = card [−2, −1]+card N = c+ℵ0 c. (iii) Note that c c + c = card ((0, 1) ∪ (1, 2)) card R = c from which the required result is immediate. (iv) Observe that ℵ0 n + ℵ0 ℵ0 + ℵ0 = ℵ0 and c n + c c + c = c, from which the results follow. Next we deﬁne multiplication of cardinal numbers. Let α and β be any cardinal numbers and select disjoint A1.3.5 Deﬁnition. sets A and B such that card A = α and card B = β. Then the product of the cardinal numbers α and β is denoted by αβ and is equal to card (A × B). 411 As in the case of addition of cardinal numbers, it is necessary, but routine, to check in Deﬁnition A1.3.5 that α β does not depend on the speciﬁc choice of the sets A and B. A1.3.6 Proposition. For any cardinal numbers α, β and γ (i) αβ = βα ; (ii) α(βγ) = (αβ)γ ; (iii) 1.α = α ; (iv) 0.α = 0; (v) α(β + γ) = αβ + αγ; (vi) For any ﬁnite cardinal n, nα = α + α + . . . α (n-terms); (vii) If α β then αγ βγ . Proof. Exercise A1.3.7 Proposition. (i) ℵ0 ℵ0 = ℵ0; (ii) c c = c; (iii) c ℵ0 = c; (iv) For any ﬁnite cardinal n, n ℵ0 = ℵ0 and n c = c. Outline Proof. (i) follows from Proposition A1.1.16, while (ii) follows from Proposition A1.2.13. To see (iii), observe that c = c.1 c ℵ0 c c = c. The proof of (iv) is also straightforward. 412 APPENDIX 1: INFINITE SETS The next step in the arithmetic of cardinal numbers is to deﬁne exponentiation of cardinal numbers; that is, if α and β are cardinal numbers then we wish to deﬁne α β. Let α and β be cardinal numbers and A and B sets A1.3.8 Deﬁnitions. such that card A = α and card B = β. The set of all functions f of B i
nto A is denoted by AB. Further, α β is deﬁned to be card AB. Once again we need to check that the deﬁnition makes sense, that is that αβ does not depend on the choice of the sets A and B. We also check that if n and m are ﬁnite cardinal numbers, A is a set with n elements and B is a set with m elements, then there are precisely nm distinct functions from B into A. We also need to address one more concern: If α is a cardinal number and A is a set such that card A = α, then we have two diﬀerent deﬁnitions of 2α. The above deﬁnition has 2α as the cardinality of the set of all functions of A into the two point set {0, 1}. On the other hand, Deﬁnition A1.2.10 deﬁnes 2α to be card (P(A)). It suﬃces to ﬁnd a bijection θ of {0, 1}A onto P(A). Let f ∈ {0, 1}A. Then f : A → {0, 1}. Deﬁne θ(f ) = f −1(1). The task of verifying that θ is a bijection is left as an exercise. For any cardinal numbers α, β and γ : A1.3.9 Proposition. (i) α β+γ = αβαγ ; (ii) (αβ)γ = αγ βγ ; γ = α(βγ) ; (iii) (αβ) (iv) α β implies αγ βγ ; (v) α β implies γα γ β . Proof. Exercise After Deﬁnition A1.2.10 we asked three questions. We are now in a position to answer the second of these questions. A1.3.10 Lemma. ℵ0 = c. ℵ0 413 Observe that card NN = ℵ0 ℵ0 and card (0, 1) = c. As the function Proof. f : (0, 1) → NN given by f (0.a1a2 . . . an . . . ) = a1, a2, . . . , an, . . . is an injection, it follows that c ℵ0 ℵ0. By the Cantor-Schröder-Bernstein Theorem A1.2.1, to conclude the proof it If a1, a2, . . . , an, . . . is any suﬃces to ﬁnd an injective map g of NN into (0, 1). element of NN, then each ai ∈ N and so we can write ai = . . . ain ai(n−1) . . . ai2 ai1, where for some Mi ∈ N, ain = 0, for all n > Mi [For example 187 = . . . 0 0 . . . 0 1 8 7 and so if ai = 187 then ai1 = 7, ai2 = 8, ai3=1 and ain = 0, for n > Mi = 3.] Then deﬁne the map g by g(a1, a2, . . . , an, . . . ) = 0.a11a12a21a13a22a31a14a23a32a41a15a24a33a42a51a16 . . . . (Compare this with the proof of Lemma A1.1.13.) Clearly g is an injection, which completes the proof. 414 APPENDIX 1: INFINITE SETS We now state a beautiful result, ﬁrst proved by Georg Cantor. A1.3.11 Theorem. 2ℵ0 = c. Firstly observe that 2ℵ0 ℵ0 ℵ0 = c, by Lemma A1.3.10. So we Proof. have to verify that c 2ℵ0. To do this it suﬃces to ﬁnd an injective map f of the set [0, 1) into {0, 1}N. Each element x of [0, 1) has a binary representation x = 0.x1x2 . . . xn . . . , with each xi equal to 0 or 1. The binary representation is unique except for representations ending in a string of 1s; for example, 1/4 = 0.0100 . . . 0 · · · = 0.0011 . . . 1 . . . . Providing that in all such cases we choose the representation with a string of zeros rather than a string of 1s, the representation of numbers in [0, 1) is unique. We deﬁne the function f : [0, 1) → {0, 1}N which maps x ∈ [0, 1) to the function f (x) : N → {0, 1} given by f (x)(n) = xn, n ∈ N. To see that f is injective, consider any x and y in [0, 1) with x = y. Then xm = ym, for some m ∈ N. So f (x)(m) = xm = ym = f (y)(m). Hence the two functions f (x) : N → {0, 1} and f (y) : N → {0, 1} are not equal. As x and y were arbitrary (unequal) elements of [0, 1), it follows that f is indeed injective, as required. A1.3.12 Corollary. αℵ0 = c. If α is a cardinal number such that 2 α c, then 415 Proof. Observe that c = 2ℵ0 αℵ0 cℵ0 = (2ℵ0)ℵ0 = 2ℵ0·ℵ0 = 2ℵ0 = c. A1.4 Ordinal Numbers9 A1.4.1 Deﬁnitions. A partially ordered set (X, ) is said to be linearly ordered (or totally ordered) if every two elements are comparable. The order is then said to be a linear order (or a total order.) The linear ordering is said to be a strict linear ordering (or a strict total ordering) if a b and b a =⇒ a = b, for a, b ∈ X. A1.4.2 Deﬁnitions. well-ordered set if every non-empty subset of S has a least element. The total ordering is then said to be a well-ordering. A strict totally ordered set (S, ) is said to be a 9For more detailed expositions of ordinal numbers see Abian [2], Ciesielski [83], Kamke [224], Roitman [344]. 416 APPENDIX 1: INFINITE SETS A1.4.3 Remark. Our next theorem says that every set can be well-ordered. While we call this a Theorem, it is equivalent to the Axiom of Choice A6.1.26 and to Zorn’s Lemma 10.2.16. Usually we start set theory with the Zermelo-Fraenkel (ZF) Axioms. (See Remark A6.1.26 and view my 3 videos on the Zermelo-Fraenkel Axioms: Video 2a – http://youtu.be/9h83ZJeiecg, Video 2b – http://youtu.be/QPSRB4Fhzko, & Video 2c – http://youtu.be/YvqUnjjQ3TQ.) In ZF, the Axiom of Choice cannot be proved. So the Well-Ordering Theorem A1.2.4 cannot be proved in ZF. If we assume any one of the Axiom of Choice, Zorn’s Lemma, or the Well-Ordering Theorem is true, then the other two can be proved. (For many equivalents of the Axiom of Choice, see Rubin and Rubin [348] and Rubin and Rubin [349].) If we add the Axiom of Choice to the Zermelo-Fraenkel Axioms, we get what is called ZFC. Many (but certainly not all) mathematicians work entirely within ZFC. A1.4.4 Theorem. set. Then there exists a well-ordering on S. [Well-Ordering Theorem] Let S be any non-empty A1.4.5 Deﬁnitions. The partially ordered sets (X, ) and (Y, ) are said to be order isomorphic (or have the same order type) if there is a bijection a b if and only if f (a) f (b). The f : X → Y such that for a, b ∈ X, function f is said to be an order isomorphism. 417 A1.4.6 Proposition. Let (X, ), (Y, ), and (Z, ) be partially ordered sets. (i) If f : X → Y is an order isomorphism, then the inverse function f ← : Y → X is also an order isomorphism. (ii) If f : X → Y and g : Y → Z are order isomorphisms, then g ◦ f : X → Z is also an order isomorphism. (iii) Let f : X → Y be an order isomorphism. If (X, ) is totally ordered, then (Y, ) is totally ordered. (iv) Let f : X → Y be an order isomorphism. If (X, ) is well-ordered, then (Y, ) is well-ordered. Proof. Exercise. A1.4.7 Remark. We have indicated that we start with the ZF axioms of set theory. Next we need to deﬁne the natural numbers. We shall use induction. We begin by deﬁning the number 0 as the empty set Ø. Then the number 1 = {0} = {Ø}; the number 2 = {0, 1} = {Ø, {Ø}}. Now using mathematical induction, we can deﬁne the number n to be {0, 1, 2, . . . , n − 1}. Further, N = {1, 2, . . . , n, . . . }. Thinking of the natural numbers as sets allows us to recognize that (i) n is well-ordered by ⊆; (ii) n /∈ n; (iii) if m ∈ n, then m /∈ m; (iv) if m ∈ n, then m ⊆ n. (v) if m ∈ n and p ∈ m, then p ∈ n. So with this in mind we see that each natural number n as well as N is a set with a natural well-ordering. This sets the stage for the deﬁnition of ordinal numbers ﬁrst given by John von Neumann (1903-1957) in his 1923 paper, von Neumann [409]. See also von Neumann [410]. 418 APPENDIX 1: INFINITE SETS A1.4.8 Deﬁnition. if it is the set of all ordinal numbers β < α well-ordered by ⊆. A set α is said to be an ordinal number (or an ordinal) A1.4.9 Proposition. A set α is an ordinal number (or an ordinal) if and only it has the following properties: (i) α is well-ordered by ⊆; (ii) α /∈ α; (iii) if β ∈ α, then β /∈ β; (iv) if β ∈ α, then β ⊆ α; (v) if β ∈ α and γ ∈ β, then γ ∈ α. Proof. Exercise A1.4.10 Remark. We see that each of the natural numbers n ∈ N, regarded as a well-ordered set, is an ordinal number. Further, the deﬁnition of the set N of all natural numbers satisﬁes Deﬁnition A1.4.8 and so is an ordinal number, which will be denoted by ω. A1.4.11 Remark. . The relation < is then clearly the same as ∈. The relation ⊆ on an ordinal number is usually denoted by We gather together in the next proposition several important results the proofs of which are straightforward. 419 A1.4.12 Proposition. (i) If α is an ordinal number and β ∈ α, then β is also an ordinal number; (ii) for ordinal numbers α, β, if α is order isomorphic to β then α = β; (iii) for ordinal numbers α, β, (a) α = β or (b) α < β or (c) β < α; (iv) if S is any non-empty set of ordinal numbers, then Si∈S Si is an ordinal number; (v) For every well-ordered set (S, ), there exists exactly one ordinal number α that is order isomorphic to (S, ). Proof. Exercise. Now we shall deﬁne the sum and product of two ordinals α and β. In case α and β are not disjoint, for the sum we replace them by the equivalent sets ({0} × α) and ({1} × β), then we deﬁne the ordering on the union of these two sets by keeping the original ordering on α and on β and making every element of α less than every element of β. A1.4.13 Deﬁnitions. denoted by α + β is the order type of the well-ordered set if α and β are ordinal numbers, then their sum, S = ({0} × α) ({1} × β) (i, δ) (j, γ) if and only if i < j or δ γ. ordered by The product, denoted by αβ is the order type of the well-ordered set α × β where the ordering is antilexicographic; that is, (a, b) (c, d) if and only if (i) b < d or (ii) b = d and a c. A1.4.14 Remark. It is readily veriﬁed that addition and multiplication of ordinal numbers is associative; that is, for ordinal numbers α, β, γ, we have α + (β + γ) = (α + β) + γ and α(βγ) = (αβ)γ. However, neither addition nor multiplication of ordinal numbers is commutative. For example it is readily proved that 3 + ω = ω = ω + 3 and ω2 = ω + ω = ω = 2ω. The following table should prove helpful. 420 APPENDIX 1: INFINITE SETS Sets Ø {0} {0, 1} {0, 1, 2} . . . {0, 1, 2, . . . , n, . . . } {0, 0, 1, 2, . . . , n, . . . } {0, 1, 2, . . . , (n − 1), 0, 1, 2, . . . , n, . . . } {0, 1, 0, 1, 0, 1, 0, 1, 0iv, . . . } {0, 1, . . . , n, 0, 1, . . . , n, 0, 1, . . . , n, 0, 1, . . . , n, . . . } {0, 1, 2, . . . , n, . . . , 0} {0, 1, 2, . . . , n, . . . , 0, 1n − 1)} {0, 1, 2, . . . , n, . . . , 0, 1, 2, . . . , n, . . . } {0, 1, . . . , n, . . . , 0, 1, . . . , n, . . . , 0, 1, . . . , n, . . . , 0iv, 1iv, . . . , niv, . . . , . . . } Ordinals 2ω = ω nω = ω2 > ω + ω > ω A1.4.15 Remark. Every set of ordinal numbers is well-ordered by . Let (S, ) be a partially ordered set. If a ∈ S, then A1.4.16 Deﬁnition. the set of all elements x ∈ S su
ch that x < a is said to be the initial segment of (S, ) determined by a. A1.4.17 Remark. Every initial segment of an ordinal number is an ordinal number. A1.4.18 Proposition. For any ordinal numbers α and β, precisely one of the following is true: (i) α = β; (ii) α is an initial segment of β; (iii) β is an initial segment of α. Proof. Exercise. The successor of an ordinal number α, denoted by A1.4.19 Deﬁnitions. α+, is the smallest ordinal number β such that β > α; that is α+ = α ∪ {α}. The ordinal γ is said to be the predecessor of α, denoted by α−, if α is the successor of γ. If α = 0 and α has no predecessor, then it is said to be a limit ordinal. 421 Clearly no ﬁnite ordinal is a limit ordinal. However ω is a A1.4.20 Remark. limit ordinal. Of course the ordinal number ω + 1 = ω+ and so ω + 1 is not a limit ordinal. Indeed for each ﬁnite ordinal n, ω + n is not a limit ordinal, but ω + ω is a limit ordinal. A1.4.21 Proposition. If Γ is a set of ordinal numbers, then γi∈Γ γi is an ordinal number and is the least upper bound of Γ that is, it is equal to sup Γ = sup γi∈Γ γi. If α is an ordinal number, then (i) γi∈α (ii) γi∈α γi = α, if α = 0 or α is a limit ordinal; γi = α−, if α = 0 and α is not a limit ordinal. Proof. Exercise. A1.4.22 Proposition. ordinal number greater than every ordinal number in Γ. For every set Γ of ordinal numbers, there exists an Proof. Exercise. A1.4.23 Remark. The class of all ordinals is not a set, it is a proper class. 422 APPENDIX 1: INFINITE SETS A1.4.24 Remark. Now we turn to the notion of exponentiation of ordinal numbers; that is, for ordinal numbers α and β we wish to deﬁne the ordinal number αβ. This has to be done with considerable care in the case of inﬁnite ordinals or the resulting set will not be well-ordered and so will not be an ordinal number. Let α and β be ordinal numbers. Let αβ be the set A1.4.25 Deﬁnition. of all functions from the set β to the set α such that only a ﬁnite number of elements of the set β map to a non-zero member of the set α. Order the functions lexicographically as follows: if f, g ∈ αβ then there exists a ﬁnite set {c1 < c2 < · · · < cn} ⊆ β such that for b ∈ β, f (b) = 0 and g(b) = 0 for b /∈ {c1, c2, . . . , cn}, Let ci be the smallest member of {c1, c2, . . . , cn} such that f (ci) = g(ci). Then deﬁne f < g if f (ci) < g(ci), otherwise deﬁne g < f . A1.4.26 Remark. It needs to be checked that αβ as deﬁned in Deﬁnition A1.4.25 is indeed an ordinal number, and in particular is a well-ordered set. Without the ﬁniteness restriction in the deﬁnition, this would not be the case. [Transﬁnite Induction] Let P (γ) be a proposition A1.4.27 Theorem. which is deﬁned for all ordinals γ. If P (0) is true and for each ordinal number α, P (β) true for all β < α imples P (α) is true, then P is true for all ordinals. If α and β are ordinal numbers, then A1.4.28 Proposition. (i) αβ = (αβ− (ii) αβ = sup γ<β ) α if β > 0 is not a limit ordinal; αγ if β is a limit ordinal and α > 0. Proof. Exercise. 423 It is readily checked that A1.4.29 Remark. (i) ω2 = ω ω; (ii) ω < ω2 < · · · < ωn, for any natural number n > 2; (iii) ωω = sup n∈ω (iv) ω < ωn < ωω, for any natural number n > 1. ωn; Each of the ordinals ω, ωω. ωωω , . . . is a countable ordinal; A1.4.30 Remark. that is, each of the sets ω, ωω. ωωω , . . . is a countable set. Further, the ordinal number ε0 is deﬁned to be sup{ω, ωω, ωωω , . . . } and it too is a countable ordinal. Clearly ωε0 = ε0. An ordinal number α which satisﬁes the equation ωα = α are said to be an ε-number. The ordinal ε0 is the smallest epsilon number. The next ordinal satisfying this equation is denoted by ε1 and equals sup{ε0 + 1, ωε0+1, ωωε0+1 . . . }. All the ε-numbers is also a countable ordinal. Now we must of necessity be vague, because otherwise we would have to include quite a lot of deep material. (See Pohlers [327].) We can continue deﬁning ordinals in a recursive way. The smallest ordinal which cannot be deﬁned recursively in terms CK. It too is of smaller ordinals is called the Church-Kleene ordinal denoted by ω1 a countable ordinal. The ﬁrst uncountable ordinal is denoted ω1. A1.4.31 Remark. How are cardinal numbers related to ordinal numbers? Every cardinal number is indeed an ordinal number.If we deﬁne Z(α) to consist of all the ordinal numbers Z(α) which are equipotent to the ordinal number α, we note that Z(α) is indeed a subset of the power set of the set α and so is a set. Further it has a smallest (or ﬁrst) element, often called an initial ordinal. Each cardinal number is the initial ordinal of some ordinal α. We see that each cardinal number is a limit ordinal ordinal. Having pointed out that ever cardinal number is an ordinal number, it is essential that we observe that cardinal arithmetic is very diﬀerent from ordinal arithmetic. One need go no further than observing that the cardinal number ℵ0 is the ordinal number ω. However, ℵℵ0 0 is uncountable, while ωω is countable. Ordinal addition and multiplication are not commutative while cardinal addition and cardinal multiplication are commutative. We conclude this section by stating Cantor’s Normal Form for ordinals. 424 APPENDIX 1: INFINITE SETS A1.4.32 Proposition. [Cantor’s Normal Form] Every ordinal number can be uniquely expressed in the form ωβ1c1 + ωβ2c2 + · · · + ωβncn, where n is a natural number, c1, c2, . . . , cn are positive integers, and β1 > β2 > . . . βn 0 are ordinal numbers. A1.5 Credits for Images 1. Galileo Galilei: Public Domain. Appendix 2: Topology Personalities René Louis Baire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 Stefan Banach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 Luitzen Egbertus Jan Brouwer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 Maurice Fréchet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Felix Hausdorﬀ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 Wacław Sierpiński . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 Credit for Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 425 426 APPENDIX 2: TOPOLOGY PERSONALITIES The source for material extracted in this appendix is primarily The MacTutor History of Mathematics Archive [390], Bourbaki [51] and http://en.wikipedia.org. In fairness all of the material in this section should be treated as being essentially direct quotes from these sources, though I have occasionally changed the words slightly, and included here only the material that I consider pertinent to this book. However, you might care to begin by reading the entertaining presentation by Barry Simon of California Institute of Technology entitled “Tales of our Forefathers” at http://tinyurl.com/hr4cq3c . If you want to learn more about the History of Mathematics, you may care to watch some or all of the YouTube videos “Math History: A course in the history of mathematics” of Associate Professor Norman J. Wildberger of The University of New South Wales, Australia at http://tinyurl.com/joelwmy . René-Louis Baire René-Louis Baire was born in Paris, France in 1874. In 1905 he was appointed to the Faculty of Science at Dijon and in 1907 was promoted to Professor of Analysis. He retired in 1925 after many years of illness, and died in 1932. Reports on his teaching vary, perhaps according to his health: “Some described his lectures as very clear, but others claimed that what he taught was so diﬃcult that it was beyond human ability to understand.” Baire 427 Stefan Banach Stefan Banach was born in Ostrowsko, Austria- Hungary – now Poland – in 1892. He lectured in mathematics at Lvov Technical University from 1920 where he completed his doctorate which is said to mark the birth of functional analysis. In his dissertation, written in 1920, he deﬁned axiomatically what today is called a Banach space. The name ’Banach space’ was coined by Fréchet. In 1924 Banach was promoted to full Professor. As well as continuing to produce a stream of important papers, he wrote textbooks in arithmetic, Banach geometry and algebra for high school. Banach’s Open Mapping Theorem of 1929 uses set-theoretic concepts which were introduced by Baire in his 1899 dissertation. The Banach-Tarski paradox appeared in a joint paper of the two mathematicians (Banach and Alfred Tarski) in 1926 in Fundamenta Mathematicae entitled Sur la décomposition des ensembles de points en partiens respectivement congruent. The puzzling paradox shows that a ball can be divided into subsets which can be ﬁtted together to make two balls each identical to the ﬁrst. The Axiom of Choice is needed to deﬁne the decomposition and the fact that it is able to give such a non-intuitive result has made some mathematicians question the use of the Axiom. The Banach-Tarski paradox was a major contribution to the work being done on axiomatic set theory around this period. In 1929, together with Hugo Dyonizy Steinhaus, he started a new journal Studia Mathematica and Banach and Steinhaus became the ﬁrst editors. The editorial policy was . . . to focus on research in functional analysis and related topics. The way that Banach worked was unconventional. He liked to do mathematical research with his colleagues in the cafés of Lvov. Stanislaw Ulam recalls frequent sessions in the Scottish Café (cf. Mauldin [278]): “It was diﬃcult to outlast or outdrink Banach during these sessions. We discussed problems proposed right there, often with no solution evident even after several hours of thinking. The next day Banach was likely to appear with several small sheets of paper containing outlines of proofs he had com
pleted.” In 428 APPENDIX 2: TOPOLOGY PERSONALITIES 1939, just before the start of World War II, Banach was elected President of the Polish Mathematical Society. The Nazi occupation of Lvov in June 1941 meant that Banach lived under very diﬃcult conditions. Towards the end of 1941 Banach worked feeding lice in a German institute dealing with infectious diseases. Feeding lice was to be his life during the remainder of the Nazi occupation of Lvov up to July 1944. Banach died in 1945. Luitzen Egbertus Jan Brouwer Luitzen Egbertus Jan Brouwer was born in 1881 in Rotterdam, The Netherlands. While an undergraduate at the University of Amsterdam he proved original results on continuous motions in four dimensional space. He obtained his Master’s degree in 1904. Brouwer’s doctoral dissertation, published in 1907, made a major contribution to the ongoing debate between Bertrand Russell and Jules Henri Poincaré on the logical foundations of mathematics. Brouwer quickly found that his philosophical ideas sparked controversy. Brouwer put a very large eﬀort into studying various problems which he attacked because they appeared on David Hilbert’s Brouwer list of problems proposed at the Paris International Congress of Mathematicians in 1900. In particular Brouwer attacked Hilbert’s ﬁfth problem concerning the theory of Lie groups. He addressed the International Congress of Mathematicians in Rome in 1908 on the topological foundations of Lie groups. Brouwer was elected to the Royal Academy of Sciences in 1912 and, in the same year, was appointed extraordinary Professor of set theory, function theory and axiomatics at the University of Amsterdam; he would hold the post until he retired in 1951. Bartel Leendert van der Waerden, who studied at Amsterdam from 1919 to 1923, wrote about Brouwer as a lecturer: Brouwer came [to the university] to give his courses but lived in Laren. He came only once a week. In general that would have not been permitted - he should have lived in Amsterdam - but for him an exception was made. ... I once interrupted him during a lecture to ask a question. Before the next week’s lesson, 429 his assistant came to me to say that Brouwer did not want questions put to him in class. He just did not want them, he was always looking at the blackboard, never towards the students. Even though his most important research contributions were in topology, Brouwer never gave courses on topology, but always on – and only on – the foundations of intuitionism. It seemed that he was no longer convinced of his results in topology because they were not correct from the point of view of intuitionism, and he judged everything he had done before, his greatest output, false according to his philosophy. As is mentioned in this quotation, Brouwer was a major contributor to the theory of topology and he is considered by many to be its founder. He did almost all his work in topology early in his career between 1909 and 1913. He discovered characterisations of topological mappings of the Cartesian plane and a number of ﬁxed point theorems. His ﬁrst ﬁxed point theorem, which showed that an orientation preserving continuous one-one mapping of the sphere to itself always ﬁxes at least one point, came out of his research on Hilbert’s ﬁfth problem. Originally proved for a 2-dimensional sphere, Brouwer later generalised the result to spheres in n dimensions. Another result of exceptional importance was proving the invariance of topological dimension. As well as proving theorems of major importance in topology, Brouwer also developed methods which have become standard tools in the subject. In particular he used simplicial approximation, which approximated continuous mappings by piecewise linear ones. He also introduced the idea of the degree of a mapping, generalised the Jordan curve theorem to n-dimensional space, and deﬁned topological spaces in 1913. Van der Waerden, in the above quote, said that Brouwer would not lecture on his own topological results since they did not ﬁt with mathematical intuitionism. In fact Brouwer is best known to many mathematicians as the founder of the doctrine of mathematical intuitionism, which views mathematics as the formulation of mental constructions that are governed by self-evident laws. His doctrine diﬀered substantially from the formalism of Hilbert and the logicism of Russell. His doctoral thesis in 1907 attacked the logical foundations of mathematics and marks the beginning of the Intuitionist School. In his 1908 paper The Unreliability of the Logical Principles Brouwer rejected in mathematical proofs the Principle of the Excluded Middle, which states that any mathematical statement is either true or false. In 1918 he published a set theory developed without using the Principle of the Excluded Middle. He was made Knight 430 APPENDIX 2: TOPOLOGY PERSONALITIES in the Order of the Dutch Lion in 1932. He was active setting up a new journal and he became a founding editor of Compositio Mathematica which began publication in 1934. During World War II Brouwer was active in helping the Dutch resistance, and in particular he supported Jewish students during this diﬃcult period. After retiring in 1951, Brouwer lectured in South Africa in 1952, and the United States and Canada in 1953. In 1962, despite being well into his 80s, he was oﬀered a post in Montana. He died in 1966 in Blaricum, The Netherlands as the result of a traﬃc accident. Maurice Fréchet Maurice Fréchet was born in France in 1878 and introduced the notions of metric space and compactness (see Chapter 7) in his dissertation in 1906. He held positions at a number of universities including the University of Paris from 1928–1948. His research includes important contributions to topology, probability, and statistics. He died in 1973. Fréchet 431 Felix Hausdorﬀ One of the outstanding mathematicians of the ﬁrst half of the twentieth century was Felix Hausdorﬀ. He did ground- breaking work in topology, metric spaces, functional analysis, Lie algebras and set theory. He was born in Breslau, Germany – now Wrocław, Poland – in 1868. He graduated from, and worked at, University of Leipzig until 1910 when he accepted a Chair at the University of Bonn. In 1935, as a Jew, he was forced to leave his academic position there by the Nazi Nuremberg Laws. He continued to do research in mathematics for several years, but could publish his results only outside Germany. In 1942 he was scheduled to go to an internment camp, but instead he and Hausdorﬀ his wife and sister committed suicide. 432 APPENDIX 2: TOPOLOGY PERSONALITIES Wacław Sierpiński Wacław Sierpiński was born in 1882 in Warsaw, Russian Empire – now Poland. Fifty years after he graduated from the University of Warsaw, Sierpiński looked back at the problems that he had as a Pole taking his degree at the time of the Russian occupation: . . . we had to attend a yearly lecture on the Russian language. . . . Each of the students made it a point of honour to have the worst results in that subject. . . . I did not answer a single question . . . and I got an unsatisfactory mark. ... I passed all my examinations, then the lector suggested I should take a repeat examination, otherwise Sierpinski I would not be able to obtain the degree of a candidate for mathematical science. . . . I refused him saying that this would be the ﬁrst case at our University that someone having excellent marks in all subjects, having the dissertation accepted and a gold medal, would not obtain the degree of a candidate for mathematical science, but a lower degree, the degree of a ‘real student’ (strangely that was what the lower degree was called) because of one lower mark in the Russian language. Sierpiński was lucky for the lector changed the mark on his Russian language course to ‘good’ so that he could take his degree. Sierpiński graduated in 1904 and worked as a school teacher of mathematics and physics in a girls’ school. However when the school closed because of a strike, Sierpiński went to Krakóv to study for his doctorate. At the Jagiellonian University in Krakóv he received his doctorate and was appointed to the University of Lvov in 1908. In 1907 Sierpiński for the ﬁrst time became interested in set theory. He happened across a theorem which stated that points in the plane could be speciﬁed with a single coordinate. He wrote to Tadeusz Banachiewicz asking him how such a result was possible. He received a one word reply (Georg) ‘Cantor’. Sierpiński began to study set theory and in 1909 he gave the ﬁrst ever lecture course devoted entirely to set theory. During the years 1908 to 1914, when he taught at the University of Lvov, he published three books in addition to many research papers. These books were The theory of Irrational numbers (1910), Outline of Set Theory (1912) and The 433 Theory of Numbers (1912). When World War I began in 1914, Sierpiński and his family happened to be in Russia. Sierpiński was interned in Viatka. However Dimitri Feddrovich Egorov and Nikolai Nikolaevich Luzin heard that he had been interned and arranged for him to be allowed to go to Moscow. Sierpiński spent the rest of the war years in Moscow working with Luzin. Together they began the study of analytic sets. When World War I ended in 1918, Sierpiński returned to Lvov. However shortly after he was accepted a post at the University of Warsaw. In 1919 he was promoted to Professor spent the rest of his life there. In 1920 Sierpiński, together with his former student Stefan Mazurkiewicz, founded the important mathematics journal Fundamenta Mathematica. Sierpiński edited the journal which specialised in papers on set theory. From this period Sierpiński worked mostly in set theory but also on point set topology and functions of a real variable. In set theory he made important contributions to the axiom of choice and to the continuum hypothesis. He studied the Sierpiński curve which describes a closed path which contains every interior point of a square –
a “space-ﬁlling curve”. The length of the curve is inﬁnity, while the area enclosed by it is 5/12 that of the square. Two fractals – Sierpiński triangle and Sierpiński carpet – are named after him. Sierpiński continued to collaborate with Luzin on investigations of analytic and projective sets. Sierpiński was also highly involved with the development of mathematics in Poland. In 1921 He had been honoured with election to the Polish Academy was made Dean of the faculty at the University of Warsaw. In 1928 he became Vice-Chairman of the Warsaw Scientiﬁc Society and, was elected Chairman of the Polish Mathematical Society. In 1939 life in Warsaw changed dramatically with the advent of World War II. Sierpiński continued working in the ‘Underground Warsaw University’ while his oﬃcial job was a clerk in the council oﬃces in Warsaw. His publications continued since he managed to send papers to Italy. Each of these papers ended with the words: The proofs of these theorems will appear in the publication of Fundamenta Mathematica which everyone understood meant ‘Poland will survive’. After the uprising of 1944 the Nazis burned his house destroying his library and personal letters. Sierpiński spoke of the tragic events of the war during a lecture he gave in 1945. He spoke of his students who had died in the war: In July 1941 one of my oldest students Stanislaw Ruziewicz was murdered. He was a retired professor of Jan Kazimierz University in Lvov . . . an outstanding mathematician and an excellent teacher. In 434 APPENDIX 2: TOPOLOGY PERSONALITIES 1943 one of my most distinguished students Stanislaw Saks was murdered. He was an Assistant Professor at Warsaw University, one of the leading experts in the world in the theory of the integral. . . In 1942 another student of mine was Adolf Lindenbaum was murdered. He was an Assistant Professor at Warsaw University and a distinguished author of works on set theory. After listing colleagues who were murdered in the war such as Juliusz Pawel Schauder and others who died as a result of the war such as Samuel Dickstein and Stanislaw Zaremba, Sierpiński continued: Thus more than half of the mathematicians who lectured in our academic schools were killed. It was a great loss for Polish mathematics which was developing favourably in some ﬁelds such as set theory and topology . . . In addition to the lamented personal losses Polish mathematics suﬀered because of German barbarity during the war, it also suﬀered material losses. They burned down Warsaw University Library which contained several thousand volumes, magazines, mathematical books and thousands of reprints of mathematical works by diﬀerent authors. Nearly all the editions of Fundamenta Mathematica (32 volumes) and ten volumes of Mathematical Monograph were completely burned. Private libraries of all the four Professors of mathematics from Warsaw University and also quite a number of manuscripts of their works and handbooks written during the war were burnt too. Sierpiński was the author of the incredible number of 724 papers and 50 books. He retired in 1960 as Professor at the University of Warsaw but he continued to give a seminar on the theory of numbers at the Polish Academy of Sciences up to 1967. He also continued his editorial work, as Editor-in-Chief of Acta Arithmetica which he began in 1958, and as an editorial board member of Rendiconti dei Circolo Matimatico di Palermo, Compositio Mathematica and Zentralblatt für Mathematik. Andrzej Rotkiewicz, who was a student of Sierpiński’s wrote: Sierpiński had exceptionally good health and a cheerful nature. . . . He could work under any conditions. Sierpiński died in 1969. Credit for Images 1. René Louis Baire. Public Domain. 2. Stefan Banach. Copyright Hussein Abdulhakim. Creative Commons Licence. https://en.wikipedia.org/wiki/Creative_Commons 3. Luitzen Egbertus Jan Brouwer. Public Domain. 4. Maurice Fréchet. Public Domain. 5. Felix Hausdorﬀ. Public Domain. 6. Wacław Sierpiński. Public Domain. 435 Appendix 3: Chaos Theory and Dynamical Systems §A3.0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 §A3.1 Iterates and Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 §A3.2 Fixed Points and Periodic Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440 §A3.3 Phase Portraits, Attracting and Repelling Fixed Points . . . . . . . . . . . . . . . . 441 §A3.4 Graphical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444 §A3.5 Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449 §A3.6 The Magic of Period 3: Period 3 Implies Chaos . . . . . . . . . . . . . . . . . . . . . . . 455 §A3.7 Chaotic Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460 §A3.8 Conjugate Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466 436 §A3.0 Introduction 437 In this Appendix we give but a taste of dynamical systems and chaos theory. Most of the material is covered by way of exercises. Some parts of this Appendix require If you have not studied calculus10 you can skip this some knowledge of calculus. Appendix altogether or merely skim through it to get a ﬂavour. A3.1 Iterates and Orbits Let S be a set and f a function mapping the set S into A3.1.1 Deﬁnition. itself; that is, f : S → S. The functions f 1, f 2, f 3, . . . , f n, . . . are inductively deﬁned as follows: f 1 : S → S is given by f 1(x) = f (x); that is is given by f 1(x) = f (f (x)); that is is given by f 3(x) = f (f (f (x))); that is; and if f n−1 is known then f n : S → S is deﬁned by f n(x) = f (f n−1(x)); that is, f n = f ◦ f n−1. Each of the the functions f 1, f 2, f 3, . . . , f n, . . . is said to be an iterate of the function f . Note that f n+m = f n ◦ f m, for n, m ∈ N. Let f be a function mapping the set S into itself. If A3.1.2 Deﬁnitions. x0 ∈ S, then the sequence x0, f 1(x0), f 2(x0), . . . , f n(x0), . . . is called the orbit of the point x0. The point x0 is called the seed of the orbit. There are several possibilities for orbits, but the most important kind is a ﬁxed point. 10If you would like to refresh your knowledge in this area, you might like to look at the classic book “A course of pure mathematics” by G.H. Hardy, which is available to download at no cost from Project Gutenberg at http://www.gutenberg.org/ebooks/38769. 438 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS A3.1.3 Deﬁnition. a ∈ S is said to be a ﬁxed point of f if f (a) = a. Let f be a mapping of a set S into itself. A point Graphically, we can ﬁnd all ﬁxed points of a function A3.1.4 Example. f : R → R, simply by sketching the curve y = f (x) and seeing where it intersects the line y = x. At points of interesection, and only for these points, do we have f (x) = x. A3.1.5 Example. 439 Exercises A3.1 1. Let the functions f : R → R, g : R → R and h : R → R be given by f (x) = x(1 − x), g(x) = x sin x, and h(x) = x2 − 2, for all x ∈ R. (a) Evaluate f 1(x) and f 2(x). (b) Evaluate g2(x) and g2(1). (c) Evaluate h2(x) and h3(x). 2. (a) If C(x) = cos(x), use your calculator [in radians to 4 decimal places] to compute C10(123), C20(123), C30(123), C40(123), C50(123), C60(123), C70(123), C80(123), C90(123), C100(123), C100(500) and C100(1). What do you notice? (b) If S(x) = sin(x), use your calculator to compute S10(123), S20(123), S30(123), S40(123), S50(123), S60(123), S70(123), S80(123), S90(123), S100(123). What do you notice? 440 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS 3. Let the function h : R → R be given by h(x) = x2, for all x ∈ R. Calculate the orbits for the function h of each of the following seeds: 0, 1, −1, 0.5, 0.25. 4. Find all the ﬁxed points of the function f in Exercise 1 above. 5. Let f : R → R be given by f (x) = x3 − 3x. Find all the ﬁxed points of the function f . A3.2 Fixed Points and Periodic Points Let f be a mapping of a set S into itself. A point A3.2.1 Deﬁnition. a ∈ S is said to be eventually ﬁxed if a is not a ﬁxed point, but some point on the orbit of a is a ﬁxed point. Let f be a function mapping the set S into itself. If A3.2.2 Deﬁnitions. x ∈ S, then the point x ∈ S is said to be periodic if there exists a positive integer p such that f p(x) = x. If m is the least n ∈ N such that f n(x) = x, then m is called the prime period of x. Let f be a function mapping the set S into itself. Then A3.2.3 Deﬁnition. the point x0 ∈ S is said to be eventually periodic if x0 is not periodic itself, but some point in the orbit of x0 is periodic. A3.2.4 Remark. We have seen that points may be ﬁxed, eventually ﬁxed, periodic, or eventually periodic. However, it is important to realize that most points are not in any of these classes. Exercises A3.2 1. Verify that the point −1 is an eventually ﬁxed point of f (x) = x2. 2. Find the eventually ﬁxed points of the function f : R → R given by f (x) = \|x\|. 441 3. If f : R → R is given by f (x) = − 3 2x2 + 5 2x + 1, verify that f (0) = 1, f (1) = 2, and f (2) = 0, so that the orbit of 0 is 0, 1, 2, 0, 1, 2, . . . . Hence 0 is a periodic point of prime period 3. 4. Prove that if x is a periodic point of prime period m of the function f : S → S, then the orbit of x has precisely m points. [Hint: Firstly write down the orbit of the point x and then deduce that it has at most m points in it. Next suppose that there are fewer than m distinct points in the orbit of x and show that this leads to a contradiction to x having period m.] 5. Let f : R → R be given by f (x) = x2 − 1. Verify that the points √ 2 and 1 are eventually periodic. 6. Consider the function f : R → R given by f (x) = \|1 − x\|. (i) Find all of the ﬁxed points of f . (ii) If m is an odd integer, what can you say about the orbit of m? (iii) If m is an
even integer, what can you say about the orbit of m? A3.3 Phase Portraits, Attracting and Repelling Fixed Points We wish to study dynamical systems, that is processes in motion. Such processes include for example the motion of planets, but other systems to which this theory is applied include the weather and population growth. Some even feel the study of dynamical systems will help us to understand stock market movements. A very good method for depicting all orbits of a dynamical system is the phase portrait of the system. This is a picture on the real line of the orbits. In the phase portrait we represent ﬁxed points by solid dots and the dynamics along orbits by arrows. 442 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS if f (x) = x3, then the ﬁxed points are 0, 1, and −1. A3.3.1 Example. If \|x0\| < 1 then the orbit of x0 is a sequence which tends to 0; we write this f n(x0) → 0. If \|x0\| > 1, then the orbit is a sequence which diverges to ∞; that is, f n(x0) → ±∞. The phase portrait is given below: Let a be a ﬁxed point of the function f : R → R. The A3.3.2 Deﬁnition. point a is said to be an attracting ﬁxed point of f if there is an open interval I containing a such that if x ∈ I, then f n(x) → a as n → ∞. Let a be a ﬁxed point of the function f : R → R. The A3.3.3 Deﬁnition. point a is said to be a repelling ﬁxed point of f if there is is an open interval I containing a such that if x ∈ I with x = a then there is an integer n such that f n(x) /∈ I. A3.3.4 Example. while −1 and 1 are repelling ﬁxed points of this function. Observe that 0 is an attracting ﬁxed point of f (x) = x3, A3.3.5 Deﬁnition. A ﬁxed point which is neither repelling nor attracting is called a neutral ﬁxed point. Exercises A3.3 443 1. Verify that the picture below is a correct phase portrait of f (x) = x2 and identify whether the ﬁxed points are repelling, attracting or neutral. 2. Do phase portraits for each of the following functions f : R → R. Determine whether any ﬁxed points are attracting, repelling or neutral. (i) f (x) = −x3. (ii) f (x) = 4x. (iii) f (x) = x − x2. (iv) f (x) = sin x. 444 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS 3. Let D : [0, 1) → R be the doubling function deﬁned by D(x) = 2x, 2x − 1. 2 [We could deﬁne D more succinctly by D(x) = 2x (mod 1).] (i) Verify that the point 1 (ii) Explain why 1 99 is a periodic point and ﬁnd its prime period. n is either a periodic point or an eventually periodic point for each positive integer n. (iii) Explain why 1 2n is eventually ﬁxed, for every positive integer n. (iv) Write an explicit formula for D2(x) and D3(x), for 0 x < 1. (v) Find all ﬁxed points of D2 and D3. 4. Do a phase portrait of the function f (x) = 2x(1 − x). [This is an example of a so-called logistic function which arises naturally in the study of population growth and ecology.] A3.4 Graphical Analysis A3.4.1 Remark. We have used phase portraits to determine whether a point x0 is ﬁxed, periodic, eventually periodic etc. This method is particularly useful when we are dealing with more than one dimension. But for functions f : R → R, we can use graphical analysis. This is done as follows. Given a function f : R → R, we are asked to determine the nature of the point x0 ∈ R. What we do is ﬁnd the orbits of points a near to x0. We begin by sketching the function f and superimposing on its graph the graph of the line y = x. To ﬁnd the orbit of the point a, plot the point (a, a). Next draw a vertical line to meet the graph of f at the point (a, f (a)). Then draw a horizontal line to meet the line y = x at the point f (a), f (a). Now draw a vertical line to meet the graph of f at the point (f (a), f 2(a)). Once again draw a horizontal line to meet the line y = x at the point (f 2(a), f 2(a)). We continue this process and the points a, f (a), f 2(a), f 3(a), . . . form the orbit of a. 445 We will now consider the function f : R → R, given by A3.4.2 Example. f (x) = x4. We sketch the curve y = x4 and superimpose the line y = x. To ﬁnd the ﬁxed points we can solve f (x) = x; that is, solve x4 = x. It is readily seen that the ﬁxed points are 0 and 1. We will consider points near each of these and do a graphical analysis, as described above, to ﬁnd the orbits of points near 0 and near 1. The analysis in the diagram below shows what happens to points near to 1. The next examples show graphical analyses of two more functions to indicate how diﬀerent these can be for diﬀerent functions. You will then get experience doing graphical analysis yourself. 446 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS A3.4.3 Example. In the above ﬁgure f (x) = sin x + x + 2. A3.4.4 Example. In the above ﬁgure f (x) = x2 − 1.5. 447 448 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS Exercises A3.4 1. Determine by graphical analysis whether each ﬁxed point of f (x) = x4 is an attracting ﬁxed point, a repelling ﬁxed point, or a neutral ﬁxed point. 2. Use graphical analysis to describe the orbits of the function f (x) = 2x and to determine the type of ﬁxed point it has. 3. Find the ﬁxed points of the function f (x) = x and use graphical analysis to determine their nature (that is, whether they are attracting, repelling, or neutral √ ﬁxed points). 4. Use graphical analysis to describe the fate of all orbits of the function f (x) = x − x2. 5. Use graphical analysis to describe the fate of all orbits of the function f (x) = ex. 6. Let f (x) = \|x − 2\|. Use graphical analysis to display a variety of orbits of f . It may help to use diﬀerent colours; for example, one colour for periodic orbits, another colour for eventually periodic orbits and yet another for eventually ﬁxed orbits. 7. Let D : [0, 1) → [0, 1) be the doubling function given by D(x) = 2x mod(1). (i) Prove that x ∈ [0, 1) is a rational number if and only if x is either a periodic point or an eventually periodic point of D. (ii) Verify that the set of all periodic points of D is P = ∞ 0, n=1 1 2n − 1 , 2 2n − 1 , 3 2n − 1 , . . . , 2n − 2 2n − 1 . [Hint. It may be helpful to write down a formula for Dn and to calculate the points of intersection of the graph of Dn with the line y = x.] (iii) Verify that the set of periodic points of D is dense in [0, 1). [We shall see that this is one of the two conditions required to show that the dynamical system ([0, 1), D) is chaotic.] A3.5 Bifurcation 449 It is natural to ask if every continuous function f : S → R A3.5.1 Remark. has a ﬁxed point, where S ⊆ R? The answer is easily seen to be no . For example, if f : R → R is given by f (x) = x + 1, then obviously there are no ﬁxed points. Therefore it is remarkable that we can guarantee the existence of ﬁxed points of all continuous functions of [0, 1] into itself. More precisely, we have already seen and proved the following corollary: 5.2.11 Corollary. mapping of [0, 1] into [0, 1]. Then there exists a z ∈ [0, 1] such that f (z) = z. (Fixed Point Theorem) Let f be a continuous Of course the above corollary does not help us to ﬁnd the ﬁxed point, rather it tells us only that at least one ﬁxed point exists. It would also be nice to have a simple way of establishing whether a particular ﬁxed point is attracting, repelling, or neutral. For well-behaved functions Theorems A3.5.2 and A3.5.3 will be very useful in this regard. 450 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS Let S be an interval in R and a be a point in the interior A3.5.2 Theorem. of S. Further, let a be a ﬁxed point of a function f : S → R. If f is diﬀerentiable at the point a and \|f (a)\| < 1, then a is an attracting ﬁxed point of f . As \|f (a)\| < 1, we have \|f (a)\| < k < 1, where k is the postive real Proof. number given by k = \|f (a)\|+1 . By deﬁnition, f (a) = lim x→a 2 \| f (x)−f (a) x−a δ > 0, such that \|f (x)−f (a) f (x)−f (a) x−a . So for x “close enough” to a, we have \| k; more precisely, there exists an interval I = [a − δ, a + δ], for some x−a Since a is a ﬁxed point, f (a) = a. So \| k, for all x ∈ I with x = a. \|f (x) − a\| k\|x − a\| , for all x ∈ I. (1) This implies that f (x) is closer to a than x is, and so f (x) is in I too. So we can repeat the same argument with f (x) replacing x and obtain \|f 2(x) − a\| k\|f (x) − a\| , for all x ∈ I. From (1) and (2), we obtain \|f 2(x) − a\| k2\|x − a\| , for all x ∈ I. (2) (3) Noting that \|k\| < 1 implies that k2 < 1, we can repeat the argument again. By mathematical induction we obtain, \|f n(x) − a\| kn\|x − a\| , kn = 0. By (4) this implies that f n(x) → a as n → ∞. And we for all x ∈ I and n ∈ N. As \|k\| < 1, have proved that a is an attracting ﬁxed point. lim n→∞ (4) The proof of Theorem A3.5.3 is analogous to that of Theorem A3.5.2 and so is left as an exercise. Let S be an interval in R and a an interior point of S. A3.5.3 Theorem. Further, let a be a ﬁxed point of a function f : S → R. If f is diﬀerentiable at the point a and \|f (a)\| > 1, then a is a repelling ﬁxed point of f . A3.5.4 Remark. It is important to note that Theorem A3.5.2 and Theorem 451 A3.5.3 do not give necessary and suﬃcient conditions. Rather they say that if f exists and \|f (x)\| < 1 in an interval containing the ﬁxed point a, then a is \|f (x)\| > 1 in an interval containing the ﬁxed an attracting ﬁxed point; and if point a, then a is a repelling ﬁxed point. If neither of these conditions is true we can say nothing! For example, it is possible that f is not diﬀerentiable at a but f still has an attracting ﬁxed point at a. (This is the case, for example for f (x) = x2 −x2 for x ∈ Q for x ∈ R \ Q, which has 0 as an attracting ﬁxed point.) Even if f is diﬀerentiable at a, Theorems A3.1.17 and A3.1.18 tell us absolutely nothing if f (a) = 1. Consider f (x) = sin x. This function is diﬀerentiable at 0 with f (0) = cos(0) = 1. So Theorems A3.1.17 and A3.1.18 tell us nothing. However, 0 is an attracting ﬁxed point of f . A3.5.5 Remark. One of the most important family of functions in this theory is the family of quadratic maps Qc : R → R, where c ∈ R, and Qc(x) = x2 + c. For each diﬀerent value of c we get a diﬀerent quadratic fun
ction. But the surprising feature is that the dynamics of Qc changes as c changes. The following theorem indicates this. We leave the proof of the theorem as an exercise. 452 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS A3.5.6 Theorem. quadratic function for c ∈ R. (The First Bifurcation Theorem) Let Qc be the (i) If c > 1 4 , then all orbits tend to inﬁnity; that is, for all x ∈ R, (Qc)n(x) → ∞ as n → ∞. (ii) If c = 1 4 , then Qc has precisely one ﬁxed point at x = 1 2 and this is a neutral ﬁxed point. (iii) If c < 1 a− = 1 2(1 − √ 1 − 4c). 4 , then Qc has two ﬁxed points a+ = 1 2(1 + √ 1 − 4c) and (a) The point a+ is always repelling. (b) If −3 (c) If c < − 3 4, then a− is repelling. 4 < c < 1 4 , then a− is attracting. A3.5.7 Remark. in the above theorem that for c > 1 precisely one ﬁxed point; but for c < 1 and one at a−. We will say more about bifurcation presently. The term bifurcation means a division into two. We see 4 there are no ﬁxed points; for c = 1 4 there is 4 this ﬁxed point splits into two — one at a+ Let f be a function mapping the set S into itself. If the A3.5.8 Deﬁnition. point x ∈ S has prime period m, then the orbit of x is {x, f (x), . . . , f m−1(x)} and the orbit is called an m-cycle. Let a be a periodic point of a function f : S → S of A3.5.9 Deﬁnitions. prime period m, for some m ∈ N. [So a is clearly a ﬁxed point of f m : S → S.] Then a is said to be an attracting periodic point of f if it is an attracting ﬁxed point of f m. Similarly a is said to be a repelling periodic point of f if it is a repelling ﬁxed point of f nm. The following theorem is left as an exercise. 453 A3.5.10 Theorem. the quadratic function for c ∈ R. (The Second Bifurcation Theorem) Let Qc be 4, then Qc has no 2-cycles. (a) If − 3 (b) If − 5 q+ = 1 2(−1 + (c) If c < − 5 √ 4 , then Qc has an attracting 2-cycle, {q−, q+}, where −4c − 3) and q− = 1 2(−1 − √ −4c − 3). 4, then Qc has a repelling 2-cycle {q−, q+}. A3.5.11 Remark. In The Second Bifurcation Theorem we saw a new kind of bifurcation called a period doubling bifurcation. As c decreases below − 3 4 , two things happen: the ﬁxed point a− changes from attracting to repelling and a new 4, we have q− = q+ = − 1 2-cycle, {q−, q+} appears. Note that when c = − 3 2 = a−. So these two new periodic points originated at a− when c = − 3 4. We will have more to say about period doubling bifurcations when we consider one-parameter families of functions (such as Qc : R → R, which depends on the parameter c, and the logistic functions fλ(x) = λx(1 − x), which depend on the parameter λ). Exercises A3.5 1. Prove Theorem A3.5.3. 2. Using Theorems A3.5.2 and A3.5.3 determine the nature of the ﬁxed points of each of the following functions: (i) f1(x) = 3x. (ii) f2(x) = 1 4x. (iii) f3(x) = x3. 3. Prove The First Bifurcation Theorem A3.5.6. 454 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS 4. Let x be a periodic point of period 2 of the quadratic map Qc. Prove that (i) Prove that x4 + 2cx2 − x + c2 + c = 0. (ii) Why do the points a+ = 1 √ 2(1 + 1 − 4c) and a− = 1 2(1 − √ 1 − 4c) satisfy the equation in (i)? [Hint. Use The First Bifurcation Theorem A3.5.6.] (iii) Using (ii), show that if x is a periodic point of prime period 2 of Qc, then x2 + x + c + 1 = 0. (iv) Deduce that if x is a periodic point of prime period 2, then x is one of the points q+ = 1 2(−1 + √ −4c − 3) and q− = 1 2(−1 − √ −4c − 3). (v) Deduce that Qc has a 2-cycle if and only if c < − 3 4 . [Be careful to eliminate the case c = − 3 4.] (vi) Using Theorem A3.1.17 show that the quadratic function Qc has q− and q+ c(x) dx \| = \|4x3 + 4cx\| < 1 at x = q− and as attracting periodic points if \| dQ2 x = q+. (vii) Noting that q− and q+ both satisfy the equation x2 + x + c + 1 = 0 (from (iii) and (iv) above), show that 4x3 + 4cx = 4x(x2 + c) = 4x(−1 − x) = 4(c + 1). (viii) Using (vi), (vii), and (v) show that for − 5 4 < c < −3 4 , q+ and q− are attracting periodic points of Qc. (ix) Similarly show that for c < − 5 4 , q+ and q− are repelling periodic points. (x) Deduce the Second Bifurcation Theorem A3.5.10 from what has been proved above in this exercise. A3.6 The Magic of Period 3: Period 3 Implies Chaos 455 A3.6.1 Remark. In 1964, the Soviet mathematician A.N. Sarkovskii published the paper (Sarkovskii [356]) in Russian in a Ukranian journal. There he proved a remarkable theorem which went unnoticed. In 1975 James Yorke and T-Y. Li published the paper (Yorke and Li [427]) in the American Mathematical Monthly. Even though the term “chaos” had previously been used in scientiﬁc literature, it was this paper that initiated the popularisation of the term. The main result of the paper, The Period Three Theorem, is astonishing, but is a very special case of Sarkovskii’s Theorem, proved a decade earlier. The discussion here of The Period Three Theorem is based on the presentation by Robert L. Devaney in his book (Devaney [101]). (The Period Three Theorem) Let f : R → R be a A3.6.2 Theorem. continuous function. If f has a periodic point of prime period 3, then for each n ∈ N it has a periodic point of prime period n. Proof. Exercises A3.6 #1–4. 456 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS A3.6.3 Remark. The Period Three Theorem is remarkable. But as stated earlier, a much more general result is true. It is known as Sarkovskii’s Theorem. We shall not give a proof, but simply point out that the proof is of a similar nature to that above. To state Sarkovskii’s Theorem we need to order the natural numbers is the following curious way known as Sarkovskii’s ordering of the natural numbers: 3, 5, 7, 9, . . . 2 · 3, 2 · 5, 2 · 7, . . . 22 · 3, 22 · 5, 22 · 7 . . . 23 · 3, 23 · 5, 23 · 7 . . . ... . . . , 2n, 2n−1, . . . , 23, 22, 21, 1. (Sarkovskii’s Theorem) Let f : R → R be a A3.6.4 Theorem. continuous function. If f has a periodic point of prime period n and n precedes k in Sarkovskii’s ordering of the natural numbers, then f has a periodic point of prime period k. 457 A3.6.5 Remarks. (i) Firstly observe that as 3 appears ﬁrst in Sarkovskii’s ordering of the natural numbers, Sarkovskii’s Theorem implies The Period Three Theorem. (ii) Secondly note that as the numbers of the form 2n constitute the tail of Sarkovskii’s ordering of the natural numbers, it follows that if f has only a ﬁnite number of periodic points, then they must all be of the form 2n. (iii) Thirdly note that Sarkovskii’s Theorem applies to continuous functions from R If R is replaced by other spaces the theorem may become false. into itself. However, R can be replaced by any closed interval [a, b]. To see this let f : [a, b] → [a, b] be a continuous function. Then extend f to a continuous function f : R → R by deﬁning f (x) = f (x), for x ∈ [a, b]; f (x) = f (a) if x < a; and f (x) = f (b), if x > b. Then the Theorem for f can be deduced from the Theorem for f . It is remarkable that the converse of Sarkovskii’s Theorem is also true but we shall not prove it here. See (Dunn [116]) (Converse of Sarkovskii’s Theorem) Let n ∈ N A3.6.6 Theorem. and l precede n in Sarkovskii’s ordering of the natural numbers. Then there exists a continuous function f : R → R which has a periodic point of prime period n, but no periodic point of prime period l. A3.6.7 Remark. From the Converse of Sarkovskii’s Theorem it follows, for example, that there exists a continuous function of R into itself which has a periodic point of prime period 6, and hence a periodic point of prime period of each even number, but no periodic point of odd prime period except 1. Exercises A3.6 1. Let f be a continuous mapping of an interval I into R. Using Propositions 4.3.5 and 5.2.1, prove that f (I) is an interval. 458 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS 2. Use the Weierstrass Intermediate Value Theorem 5.2.9 to prove the following result: Proposition A. Let a, b ∈ R with a < b and f : I = [a, b] → R a continuous function. If f (I) ⊇ I, prove that f has a ﬁxed point in I. [Hints. (i) Show that there exists points s, t ∈ [a, b] such that f (s) = c a s and f (t) = d b t. (ii) Put g(x) = f (x) − x and observe that g is continuous and g(s) 0 and g(t) 0. (iii) Apply the Weierstrass Intermediate Value Theorem to g.] 3. Use the Weierstrass Intermediate Value Theorem 5.2.9 to prove the following result: Propostion B. Let a, b ∈ R with a < b. Further, let f : [a, b] → R be a continuous function and f ([a, b]) ⊇ J = [c, d], for c, d ∈ R with c < d. Prove that there is a subinterval I = [s, t] of I = [a, b] such that f (I) = J . [Hints. (i) Verify that f −1({c}) and f −1({d}) are non-empty closed sets. (ii) Using (i) and Lemma 3.3.2 verify that there is a largest number s such that f (s) = c. (ii) Consider the case that there is some x > s such that f (x) = d. Verify that there is a smallest number such that t > s and f (t) = d. (iii) Suppose that there is a y ∈ [s, t] such that f (y) < c. Use the Weierstrass Intermediate Value Theorem to obtain a contradiction. (iv) Show also in a similar fashion that there is no z ∈ [s, t]. such that f (z) > d. (v) Deduce that, under the condition in (ii), f ([s, t]) = [c, d] = J , as required. (vi) Now consider the case that there is no x > s such that f (x) = d. Let s be the largest number such that f (s) = d. Clearly s < s. Let t be the smallest number such that t > s and f (t) = c. Verify that f ([s, t]) = [c, d] = J , as required.] 459 4. Let f be as in The Period Three Theorem A3.6.2. So there exists a point a in R of prime period 3. So f (a) = b, f (b) = c, and f (c) = a, where a = b, a = c, and b = c. We shall consider the case a < b < c. The other cases are similarly handled. Put I0 = [a, b] and I1 = [b, c]. (i) Using Exercise 1 above, verify that f (I0) ⊇ I1. (ii) Using Exercise 1 above again, verify that f (I1) ⊇ I0 ∪ I1. (iii) Deduce from (ii) and Proposition B above that there is a closed interval A1 ⊆ I1, such that f (A1) = I1. (iv) Noting that f (A1) = I1 ⊇ A1, use Proposition B above again to show there exists a closed interval A2 ⊆ A1 such that f
(A2) = A1. (v) Observe that A2 ⊆ A1 ⊆ I1 and f 2(A2) = I1. (vi) Use mathematical induction to show that for n 3 there are closed intervals A1, A2, . . . , An−2 such that An−2 ⊆ An−3 ⊆ · · · ⊆ A2 ⊆ A1 ⊆ I1 such that f (Ai) = Ai−1, i = 2, . . . , n − 2, and f (A1) = I1. (vii) Deduce from (vi) that f n−2(An−2) = I1 and An−2 ⊆ I1. (viii) Noting that f (I0) ⊇ I1 ⊇ An−2, show that there is a closed interval An−1 ⊆ I0 such that f (An−1) = An−2. (ix) Finally, using the fact that f (I1) ⊃ I0 ⊇ An−1, show that there is a closed interval An ⊂ I1 such that f (An) = An−1. (x) Putting the above parts together we see f −→An−1 f f −→A1 −→ I1 An −→ . . . with f (Ai) = Ai−1 and f n(An) = I1. Use the fact that An ⊂ I1 and Proposition A to show that there is a point x0 ∈ An such that f n(x0) = x0. (xi) Observe from (x) that the point x0 is a periodic point of f of period n. [We have yet to show that x0 is of prime period n.] (xii) Using the fact that f (x0) ∈ An−1 ⊆ I0 and f i(x0) ∈ I1, for i = 2, . . . , n, and I0 ∩ I1 = {b}, show that x0 is of prime period n. [Note the possibility that f (x0) = b needs to be eliminated. This can be done by observing that f 3(x0) ∈ I1, but f 2(b) = a /∈ I1.] (xiii) From (xi) and (xii) and (vi), deduce that f has a periodic point of prime period n for every n 3. [We deal with the cases n = 1 and n = 2 below.] 460 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS (xiv) Use Proposition A and the fact that f (I1) ⊇ I1 to show there is a ﬁxed point of f in I1; that is there exists a periodic point of prime period 1. (xv) Note that f (I0) ⊇ I1 and f (I1) ⊇ I0. Using Proposition B show that there is a closed interval B ⊆ I0 such that f (B) = I1. Then, observe that f 2(B) ⊇ I0, and from Proposition A, deduce that there exists a point x1 ∈ B such that f 2(x1) = x1. Verify that x1 ∈ B ⊆ I0 = [a, b] while f (x1) ∈ f (B) = I1 = [b, c] and x1 = b. Deduce that x1 is a periodic point of prime period 2 of f . This completes the proof of The Period Three Theorem A3.6.2. 5. (i) Show that The Period Three Theorem 3.6.2 would be false if R were replaced by R2. [Hint. Consider a rotation about the origin.] (ii) Show that The Period Three Theorem A3.6.2 would be false if R were replaced by Rn, n 2. (iii) Show that The Period Three Theorem A3.6.2 would be false if R were replaced by S1, where S1 is the circle centred at the origin of radius 1 in R2. 6. Why is the Sarkovskii Theorem A3.6.4 true when R is replaced by the open It’s easy to deduce from [Hint. interval (a, b), for a, b ∈ R with a < b? Sarkovskii’s Theorem for R.] A3.7 Chaotic Dynamical Systems A3.7.1 Remarks. Today there are literally thousands of published research papers and hundreds of books dealing with chaotic dynamical systems. These are related to a variety of disciplines including art, biology, economics, ecology and ﬁnance. It would be folly to try to give a deﬁnitive history of chaos, a term used in the book of Genesis in the Bible and Hun-Tun (translated as chaos) in Taoism (Girardot [159]), a philospohical tradition dating back 2,200 years in China to the Han Dynasty. Here we focus on the twentieth century. It would also be folly to try here to give the “correct” deﬁnition of the mathematical concept of chaos. Rather, we shall give one reasonable deﬁnition, 461 noting there are others which are inequivalent. Indeed some mathematicians assert that no existing deﬁnition captures precisely what we want chaos to be. As stated earlier, the 1975 paper of Yorke and Li [427] triggered widespread interest in chaotic dynamical systems. However the previous year the Australian scientist Robert M. May, later Lord Robert May and President of the prestigious Royal Society of London published the paper [279] in which he stated “Some of the simplest nonlinear diﬀerence equations describing the growth of biological populations with nonoverlapping generations can exhibit a remarkable spectrum of dynamical behavior, from stable equilibrium points, to stable cyclic oscillations between 2 population points, to stable cycles with 4, 8, 16, . . . points, through to a chaotic regime in which (depending on the initial population value) cycles of any period, or even totally aperiodic but bounded population ﬂuctuations, can occur.” Jules Henri Poincaré (1854–1912), one of France’s greatest mathematicians, is acknowledged as one of the founders of a number of ﬁelds of mathematics including modern nonlinear dynamics, ergodic theory, and topology. His work laid the foundations for chaos theory. He stated in his 1903 book, a translated version of which was republished in 2003 (Poincarè [328]): “If we knew exactly the laws of nature and the situation of the universe at its initial moment, we could predict exactly the situation of that same universe at a succeeding moment. But even if it were the case that the natural laws had no longer any secret for us, we could still only know the initial situation approximately. If that enabled us to predict the succeeding situation with the same approximation, that is all we require, and we should say that the phenomenon had been predicted, that it is governed by laws. But it is not always so; it may happen that small diﬀerences in the initial conditions produce very great ones in the ﬁnal phenomena. A small error in the former will produce an enormous error in the latter. Prediction becomes impossible”. What Poincaré described quite precisely has subsequently become known colloquially as the butterﬂy eﬀect, an essential feature of chaos. 462 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS In 1952 Collier’s magazine published a short story called “A Sound of Thunder” by the renowned author, Ray Bradbury (1920–2012). In the story, http://www. lasalle.edu/~didio/courses/hon462/hon462_assets/sound_of_thunder.htm a party of rich businessmen use time travel to journey back to a prehistoric era and go on a safari to hunt dinosaurs. However, one of the hunters accidentally kills a prehistoric butterﬂy, and this innocuous event dramatically changes the future world that they left. This was perhaps the incentive for a meteorologist’s presentation in 1973 to the American Association for the Advancement of Science in Washington , D.C. being given the name “Predictability: Does the ﬂap of a butterﬂy’s wings in Brazil set oﬀ a tornado in Texas?” The meteorologist was Edward Norton Lorenz (1917–2008) and the ﬂapping wing represented a tiny change in initial conditions causing enormous changes later. Lorenz discovered sensitivity to initial conditions by accident. He was running on a computer a mathematical model to predict the weather. Having run a particular sequence, he decided to replicate it. He re-entered the number from his printout, taken part-way through the sequence, and let it run. What he found was that the new results were radically diﬀerent from his ﬁrst results. Because his printout rounded to three decimal places, he had entered the number .506 than the six digit number .506127. Even so, he would have expected that the resulting sequence would diﬀer only slightly from the original run. Since repeated experimentation proved otherwise, Lorenz concluded that the slightest diﬀerence in initial conditions made a dramatic diﬀerence to the outcome. So prediction was in fact impossible. Sensitivity to initial conditions, or the butterﬂy eﬀect, had been demonstrated to be not just of theoretical importance but in fact of practical importance in meteorology. It was a serious limitation to predicting the weather – at least with that model. Perhaps this eﬀect was evident also in a variety of other practical applications. The American mathematicians George David Birkoﬀ (1884-1944) and Harold Calvin Marston Morse (1892–1977) continued Poincaré’s work on dynamical systems. While Poincaré had made use of topology in the theory of dynamical systems, Birkhoﬀ, in particular, supplemented this by the use of Lebesgue measure theory. In 1931 Birkhoﬀ and P.A. Smith in their paper [43] introduced the concept of metric transitivity which is central in ergodic theory and was used by Robert L. Devaney in 1986 in his widely published deﬁnition of, and approach to, chaos. The three conditions: transitivity, sensitivity to initial conditions, and density of of periodic points as appeared in The Period Three Theorem, were precisely what Devaney used in his deﬁnition of chaos. 463 A3.7.2 Deﬁnition. of the set X into itself. Then (X, f ) is said to be a dynamical system. Let (X, d) be a metric space and f : X → X a mapping A3.7.3 Remark. system as (X, d, f ), however in the literature the convention is not do this. It would be much more appropriate to denote the dynamical Let (X, d) be a metric space and f : X → X a mapping A3.7.4 Deﬁnition. of X into itself. Then the dynamical system (X, f ) is said to be transitive if given x, y ∈ X, and any ε > 0, there exists a z ∈ X and an n ∈ N, such that d(z, y) < ε d(f n(z), x) < ε. and A3.7.5 Remark. “close” to y such that some point in the orbit of z is “close” to x. Roughly speaking, transitivity says that there is a point z A3.7.6 Remark. At long last we shall deﬁne chaos. However, care needs to be taken as there is a number of inequivalent deﬁnitions of chaos in the literature as well as many writers who are vague about what they mean by chaos. Our deﬁnition is that used by Robert L. Devaney, with a modiﬁcation resulting from the work of a group of Australian mathematicians, Banks et al. [31], in 1992. A3.7.7 Deﬁnition. The dynamical system (X, f ) is said to be chaotic if (i) the set of all periodic points of f is dense in the set X, and (ii) (X, f ) is transitive. 464 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS A3.7.8 Remark. Until 1992 it was natural to add a third condition in the deﬁnition of chaotic dynamical systems. This condition is that in the dynamical system (X, f ), f depends sensitively on initial conditions. However in 1992 a group of mathematicians from La Trobe University in Melbourne, Australia proved tha
t this condition is automatically true if the two conditions in Deﬁnition A3.7.7 hold. Their work appeared in the paper “On Devaney’s deﬁnition of chaos" by the authors John Banks, Gary Davis, Peter Stacey, Jeﬀ Brooks and Grant Cairns in the American Mathematical Monthly (Banks et al. [31]). See also (Banks et al. [32]). The dynamical system (X, f ) is said to depend A3.7.9 Deﬁnition. sensitively on initial conditions if there exists a β > 0 such that for any x ∈ X and any ε > 0 there exists n ∈ N and y ∈ X with d(x, y) < ε such that d(f n(x), f n(y)) > β. This deﬁnition says that no matter which x we begin A3.7.10 Remark. with and how small a neighbourhood of x we choose, there is always a y in this neighbourhood whose orbit separates from that of x by at least β. [And β is independent of x.] A3.7.11 Remark. What we said in Remark A3.7.8 is that every chaotic dynamical system depends sensitively on initial conditions. We shall not prove this here. But we will show in Exercises A3.7 #2 that the doubling map does indeed depend sensitively on initial conditions. A3.7.12 Remark. In 1994, Michel Vellekoop and Raoul Berglund [402] proved that in the special case that (X, d) is a ﬁnite or inﬁnite interval with the Euclidean metric, then transitivity implies condition (ii) in Deﬁnition A3.7.7, namely that the set of all periodic points is dense. However, David Asaf and Steve Gadbois [213] showed this is not true for general metric spaces. Exercises A3.7 465 1. Let D : [0, 1) → [0, 1) given by D(x) = 2x (mod 1) be the doubling map. Prove that the dynamical system ([0, 1), D) is chaotic. [Hints. Recall that in Exercises A3.4 #7 it was proved that the set of all periodic points of D is ∞ 0, 1 2n − 1 , 2 2n − 1 , 3 2n − 1 , . . . , 2n − 2 2n − 1 . P = n=1 and that the set P is dense in [0, 1). So condition (i) of Deﬁnition A3.7.7 is satisﬁed. To verify condition (ii) use the following steps: (a) Let x, y ∈ [0, 1) and ε > 0 be given. Let n ∈ N be such that 2−n < ε. For k ∈ {1, 2, . . . , n}, let Jk,n = k − 1 2n , k 2n . Show that there exists a k ∈ {1, 2, . . . , n}, such that x ∈ Jk,n. (b) Verify that f n(Jk,n) = [0, 1). (c) Deduce from (b) that there exists a z ∈ Jk,n such that f n(z) = y. (d) Deduce that z has the required properties of Deﬁnition A3.7.4 and so ([0, 1), D) is a transitive dynamical system. (e) Deduce that ([0, 1), D) is a chaotic dynamical system.] 2. Prove that the doubling map of Exercise 1 above depends sensitively on initial conditions. [Hints. Let β = 1 4. Given any ε > 0, let n ∈ N be such that 2−n < ε. Put s = f n(x) + 0.251 (mod 1). Firstly, verify that d(f n(x), s) > β. As observed in Exercise 1(a), x ∈ Jk,n, for some k ∈ {1, 2, . . . , n}. But by Exercise 1(b), f n(Jk,n) = [0, 1). Let y ∈ Jk,n be such that f n(y) = s. Now verify that y has the required properties (i) d(x, y) < ε and (ii) d(f n(x), f n(y) > β.] 3. Let m be a (ﬁxed) positive integer and consider the dynamical system ([0, 1), fm) where fm(x) = mx (mod 1). Prove that ([0, 1), fm) is chaotic. [Hint. See Exercise 1 above.] 466 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS 4. Let (X, τ ) be a topological space and f a continuous mapping of X into itself. Then f is said to be topologically transitive if for any pair of non-empty open sets U and V in (X, τ ) there exists an n ∈ N such that f k(U ) ∩ V = Ø. For X, d) a metric space and τ the topology induced by the metric d, prove that f is transitive if and only if it is topologically transitive. A3.8 Conjugate Dynamical Systems Let (X1, d1) and (X2, d2) be metric spaces and (X1, f1) A3.8.1 Deﬁnition. and (X2, f2) dynamical systems. Then (X1, f1) and (X2, f2) are said to be conjugate dynamical systems if there is a homeomorphism h : (X1, d1) → (X, d2) such that f2 ◦ h = h ◦ f1; that is, f2(h(x)) = h(f1(x)), for all x ∈ X1. The map h is called a conjugate map. In Exercises A3.8 #2 it is verﬁed that if (X1, f1) and (X2, f2) 3A.8.2 Remark. are conjugate dynamical systems, then (X2, f2) and (X1, f1) are conjugate dynamical systems. So we see that the order in which the dynamical systems are considered is of no importance. A3.8.3 Remark. Conjugate dynamical systems are equivalent in the same sense that homeomorphic topological spaces are equivalent. The next theorem demonstrates this fact. Very often it will be possible to analyze a complex dynamical system by showing it is conjugate to one we already understand. 467 A3.8.4 Theorem. systems, where h is the conjugate map. (i) A point x ∈ X1 is a ﬁxed point of f1 in X1 if and only if h(x) is a ﬁxed Let (X1, f1) and (X2, f2) be conjuagate dynamical point of f2 in X2. (ii) A point x ∈ X1 is a periodic point of period n ∈ N of f1 in X1 if and only if h(x) is a periodic point of period n of f2 in X2. (iii) The dynamical system (X, f1) is chaotic if and only if the dynamical system (X, f2) is chaotic. Proof. (i) and (ii) are straightforward and left as exercises for you. To see (iii), assume that (X1, f1) is chaotic. Let P be the set of periodic points of f1. As (X1, f1) is chaotic, P is dense in X1. As h is a continuous, it is easily seen that h(P ) is dense in the set h(X1) = X2. As h(P ) is the set of periodic points of (X2, f2), it follows (X2, f2) satisﬁes condition (i) of Deﬁnition A3.7.7. To complete the proof, we need to show that (X2, f2) is transitive. To this end, let ε > 0 and u, v ∈ X2. Then there are x, y ∈ X1 such that h(x) = u and h(y) = v. Since h is continuous, it is continuous at the points x, y ∈ X1. Thus, there exists a δ > 0 such that z ∈ X1 and d1(x, z) < δ ⇒ d2(h(x), h(z)) < ε, and z ∈ X1 and d1(y, z) < δ ⇒ d2(h(y), h(z)) < ε. As (X1, f1) is transitive, there is a z ∈ X1 and n ∈ N, such that d1(x, z) < δ ⇒ d1(f n 1 (z), y) < δ. (12.1) (12.2) (12.3) Let z be chosen so that (12.3) holds, and put w = h(z). Using this value for z in (12.1), and taking f n 1 (z) as z in (12.2), we obtain d2(u, w) = d2(h(x), h(z)) < ε, from (12.1) and (12.3) (12.4) and d2(f n 2 (w), v) = d2(f n 2 (h(z)), h(y)), 1 (z), h(y)), = d2(h(f n < ε, as h ◦ f1 = f2 ◦ h, (12.5) from (2) and (3). Now from (12.4) and (12.5) it follows that (X2, f2) is transitive. 468 APPENDIX 3: CHAOS THEORY AND DYNAMICAL SYSTEMS Exercises A3.8 1. Let T : [0, 1] → [0, 1] be the tent function given by T (x) = 2x, 2 − 2x, for 0 x 1 2, for 1 x 1. 2 (i) Sketch the graph of T . (ii) Calculate the formula for T 2 and sketch the graph of T 2. (iii) Calculate the formula for T 3 and sketch the graph of T 3. (iv) Let Ik,n = [ k−1 T n(Ik,n) = [0, 1]. 2n , k 2n ], for k ∈ {1, 2, . . . , 2n − 1}, n ∈ N. Verify that (v) Using Proposition A of Exercises 3.6 #2, show that T n has a ﬁxed point in each Ik,n. (vi) Deduce from (v) that there is a periodic point of T in each Ik,n. (vii) Using the above results show that (T, [0, 1]) is a chaotic dynamical system. 2. Verify that if (X1, f1) and (X2, f2) are conjugate dynamical systems then (X2, f2) (So the order in which the and (X1, f1) are conjugate dynamical systems. dynamical systems are considered is of no importance.) 3. Let L : [0, 1] → [0, 1] be the logistic function given by L(x) = 4x(1 − x). (i) Show that the map h : [0, 1] → [0, 1] given by h(x) = sin2( π is a homeomorphism of [0, 1] onto itself such that h ◦ T = L ◦ h, where T is the tent function. 2 x), (ii) Deduce that ([0, 1], T ) and ([0, 1], L) are conjugate dynamical systems. (iii) Deduce from (ii), Theorem A3.8.4 and Exercise 1 above that ([0, 1], L) is a chaotic dynamical system. 4. Consider the quadratic map Q−2 : [−2, 2] → [−2, 2], where Q−2(x) = x2 − 2. (i) Prove that the dynamical systems ([−2, 2], Q−2) and ([0, 1], L) of Exercise 3 above are conjugate. (ii) Deduce that ([−2, 2], Q−2) is a chaotic dynamical system. Appendix 4: Hausdorﬀ Dimension §A4.0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 §A4.1 Hausdorﬀ Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 469 470 APPENDIX 4: HAUSDORFF DIMENSION §A4.0 Introduction In this section we introduce the notion of Hausdorﬀ Dimension which plays an important role in the study of fractals. A4.1 Hausdorﬀ Dimension We begin by warning the reader that this section is signiﬁcantly more complicated than much of the material in the early chapters of this book. Further, an understanding of this section is not essential to the understanding of the rest of the book. We think of points as 0-dimensional, lines as 1-dimensional, squares as 2- dimensional, cubes as 3-dimensional etc. So intuitively we think we know what the notion of dimension is. For arbitrary topological spaces there are competing notions In “nice” spaces, the diﬀerent notions of topological of topological dimension. dimension tend to coincide. However, even the well-behaved euclidean spaces, Rn, n > 1, have surprises in store for us. In 1919 Felix Hausdorﬀ [172] introduced the notion of Hausdorﬀ dimension of a metric space. A surprising feature of Hausdorﬀ dimension is that it can have values which are not integers. This topic was developed by Abram Samoilovitch Besicovitch [40] a decade or so later, but came into prominence in the 1970s with the work of Benoit Mandelbrot on what he called fractal geometry and which spurred the development of chaos theory. Fractals and chaos theory have been used in a very wide range of disciplines including economics, ﬁnance, meteorology, physics, and physiology. We begin with a discussion of Hausdorﬀ measure (or what some call Hausdorﬀ- Besicovitch measure). Some readers will be familiar with the related notion of Lebesgue measure, however such an understanding is not essential here. Let Y be a subset of a metric space (X, d). Then the A4.1.1 Deﬁnition. number sup{d(x, y) : x, y ∈ Y } is said to be the diameter of the set Y and is denoted diam Y . Let Y be a subset of a metric space (X, d), I an index A4.1.2 Deﬁnition. set, ε a positive real number, and {Ui : i ∈ I} a family of subsets of X such t
hat Y ⊆ Ui and, for each i ∈ I, diam Ui < ε. Then {Ui : i ∈ I} is said to i∈I be an ε-covering of the set Y . 471 We are particularly interested in ε-coverings which are countable. So we are led to ask: which subsets of a metric space have countable ε-coverings for all ε > 0? The next Proposition provides the answer. Let Y be a subset of a metric space (X, d) and d1 A4.1.3 Proposition. the induced metric on Y . Then Y has a countable ε-covering for all ε > 0 if and only if (Y, d1) is separable. Assume that Y has a countable ε-covering for all ε > 0. In particular Y Proof. has a countable (1/n)-covering, {Un,i : i ∈ N}, for each n ∈ N. Let yn,i be any point in Y ∩ Un,i. We shall see that the countable set {yn,i : i ∈ N, n ∈ N} is dense in Y. Clearly for each y ∈ Y , there exists an i ∈ N, such that d(y, yn,i) < 1/n. So let O be any open set intersecting Y non-trivially. Let y ∈ O ∩ Y . Then O contains an open ball B centre y of radius 1/n, for some n ∈ N. So yn,i ∈ O, for some i ∈ N. Thus {yn,i : i ∈ N, n ∈ N} is dense in Y and so Y is separable. Conversely, assume that Y is separable. Then it has a countable dense subset {yi : i ∈ N}. Indeed, given any y ∈ Y and any ε > 0, there exists a yi, i ∈ N, such that d(y, yi) < ε/2. So the family of all {Ui : i ∈ N}, where Ui is the open ball centre yi and radius ε/2 is an ε-covering of Y , as required. We are now able to deﬁne the Hausdorﬀ s-dimensional measure of a subset of a metric space. More precisely, we shall deﬁne this measure for separable subsets of a metric space. Of course, if (X, d) is a separable metric space, such as Rn, for any n ∈ N, then all of its subsets are separable. (See Exercises 6.3 #15.) 472 APPENDIX 4: HAUSDORFF DIMENSION A4.1.4 Deﬁnition. and s a positive real number. For each positive real number ε < 1, put Let Y be a separable subset of a metric space (X, d) Hs ε(Y ) = inf (diam Ui)s : {Ui : i ∈ N} is an ε-covering of Y , and Hs(Y ) = i∈N Hs ε(Y ),    lim ε→0 ε>0 ∞, if the limit exists; otherwise. Then Hs(Y ) is said to be the s-dimensional Hausdorﬀ outer measure of the set Y . A4.1.5 Remark. Hs ε2 helps us to understand the deﬁnition of Hs(Y ). (Y ). So as ε tends to 0, either the limit of Hs (Y ) ε(Y ) exists or it tends to ∞. This Note that in Deﬁnition A4.1.4, if ε1 < ε2, then Hs ε1 It is important to note that if d1 is the metric induced on Y A4.1.6 Remark. by the metric d on X, then Hs(Y ) depends only on the metric d1 on Y . In other words if Y is also a subset of the metric space (Z, d2) and d2 induces the same metric d1 on Y , then Hs(Y ) is the same when considered as a subset of (X, d) or (Y, d2). So, for example, the s-dimensional Hausdorﬀ outer measure is the same for the closed interval [0,1] whether considered as a subset of R or of R2 or indeed of Rn, for any positive integer n. Let Y be a separable subset of a metric space (X, d), s A4.1.7 Lemma. and t positive real numbers with s < t, and ε a positive real number < 1. Then 473 (i) Ht (ii) Ht ε (Y ), and ε(Y ) Hs ε(Y ) εt−sHs ε (Y ). Part (i) is an immediate consequence of the fact that ε < 1 and so each Proof. diam Ui < 1, which implies (diam Ui)t < (diam Ui)s. Part (ii) follows from the fact that diam Ui < ε < 1 and so (diam Ui)t < εt−s(diam Ui)s. A4.1.8 Proposition. and s and t positive real numbers with s < t. Let Y be a separable subset of a metric space (X, d) (i) If Hs(Y ) < ∞, then Ht(Y ) = 0. (ii) If 0 = Ht(Y ) < ∞, then Hs(Y ) = ∞. Proof. These follow immediately from Deﬁnition A4.1.3 and Lemma A4.1.7ii). From Proposition A4.1.8 we see that if Hs(Y ) is ﬁnite and A4.1.9 Remark. non-zero for some value of s, then for all larger values of s, Hs(Y ) equals 0 and for all smaller values of s, Hs(Y ) equals ∞. Proposition A4.1.8 allows us to deﬁne Hausdorﬀ dimension. A4.1.10 Deﬁnition. Let Y be a separable subset of a metric space (X, d). Then dimH(Y ) = inf{s ∈ [0, ∞) : Hs(Y ) = 0}, ∞, if Hs(Y ) = 0 for some s > 0; otherwise is called the Hausdorﬀ dimension of the set Y . 474 APPENDIX 4: HAUSDORFF DIMENSION We immediately obtain the following Proposition. A4.1.11 Proposition. (X, d). Then (i) (ii) dimH(Y ) = Hs(Y ) = Let Y be a separable subset of a metric space   0, sup{s ∈ [0, ∞) : Hs(Y ) = ∞},  ∞, 0, ∞, if s > dimH(Y ); if s < dimH(Y ). if Hs(Y ) = 0 for all s; if the supremum exists; otherwise. The calculation of the Hausdorﬀ dimension of a metric space is not an easy exercise. But here is an instructive example. A4.1.12 Example. dimH(Y ) = 0. Let Y be any ﬁnite subset of a metric space (X, d). Then Put Y = {y1, y2, . . . , yN }, N ∈ N. Let Oε(i) be the open ball centre yi Proof. and radius ε/2. Then {Oi : i = 1, . . . , N } is an ε-covering of Y . So Hs ε(Y ) = inf (diam Ui)s : {Ui} an open covering of Y } i∈N N (diam Oi)s = εs.N s+1/2s. i=1 εs.N s+1/2s = 0. So Hs(Y ) = 0, for all s > 0. Hence Thus Hs(Y ) lim ε→0 ε>0 dimH(Y ) = 0. The next Proposition is immediate. A4.1.13 Proposition. If (Y1, d1) and (Y2, d2) are isometric metric spaces, then dimH(Y1) = dimH(Y2). A4.1.14 Proposition. (X, d). If Z ⊂ Y , then dimH(Z) dimH(Y ). Let Z and Y be separable subsets of a metric space 475 Proof. Exercise. A4.1.15 Lemma. (X, d). Then Let Y = i∈N Yi be a separable subset of a metric space Hs(Y ) ∞ i=1 Hs(Yi). Proof. Exercise. A4.1.16 Proposition. space (X, d). Then Let Y = i∈N Yi be a separable subset of a metric dimH(Y ) = sup{dimH(Yi) : i ∈ N}. Proof. It follows immediately from Lemma A4.1.15 that dimH(Y ) sup{dimH(Yi) : i ∈ N}. However, by Proposition A4.1.14, dimH(Y ) dimH(Yi), for each i ∈ N. Putting these two observations together completes the proof of the Proposition. 476 APPENDIX 4: HAUSDORFF DIMENSION A4.1.17 Proposition. then dimH(Y ) = 0. If Y is a countable subset of a metric space (X, d), Proof. This follows immediately from Proposition A4.1.16 and Example A4.1.12 In particular, Proposition A4.1.17 tells us that dimH(Q) = 0. A4.1.18 Example. the euclidean metric. Then dimH[a, a + 1] = dimH[0, 1] = dimH(R). Let [a, a + 1], a ∈ R be a closed interval in R, where R has Let da be the metric induced on [a, a + 1] by the euclidean metric on Proof. R. Then ([a, a + 1], da) is isometric to ([0, 1], d0), and so by Proposition A4.1.13, dimH[a, a + 1] = dimH[0, 1]. Now observe that R = [a, a + 1]. So ∞ a=−∞ dimH(R) = sup{dimH[a, a+1] : a = . . . , −n, . . . , −2, −1, 0, 1, 2, . . . , n, . . . } = dimH[0, 1], as each dimH[a, a + 1] = dimH[0, 1]. Let (X, d1) and (Y, d2) be separable metric spaces A4.1.19 Proposition. and f : X → Y a surjective function. If there exist positive real numbers a and b, such that for all x1, x2 ∈ X, a.d1(x1, x2) d2(f (x1), f (x2)) b.d1(x1, x2), then dimH(X, d1) = dimH(Y, d2). Proof. Exercise A4.1.20 Remark. In some cases Proposition A4.1.19 is useful in calculating the Hausdorﬀ dimension of a space. See Exercises 6.6 #7 and #8. Another useful device in calculating Hausdorﬀ dimension is to reﬁne the deﬁnition of the s-dimensional Hausdorﬀ outer measure as in the following Proposition, where all members of the ε-covering are open sets. 477 A4.1.21 Proposition. and s a positive real number. If for each positive real number ε < 1, Let Y be a separable subset of a metric space (X, d) Os ε(Y ) = inf (diam Oi)s : {Oi : i ∈ N} is an ε-covering of Y by open sets Oi , then Os ε(Y ) = Hs   i∈N ε(Y ). lim ε→0 ε>0 ∞,  Os ε(Y ), Further Hs(Y ) = if the limit exists; otherwise. Proof. Exercise. A4.1.22 Lemma. (X, d). If {Oi : i ∈ N} is a covering of Y by open sets Oi, then Let Y be a connected separable subset of a metric space diam Oi diam Y i∈N Proof. Exercise. A4.1.23 Example. Show H1[0, 1] 1. Proof. noting diam[0, 1] = 1 yields H1 result. If we put Y = [0, 1] in Lemma A4.1.22 and s = 1 in Proposition A4.1.21, ε[0, 1] 1, for all ε > 0. This implies the required 478 APPENDIX 4: HAUSDORFF DIMENSION A4.1.24 Proposition. euclidean metric. Then dimH[0, 1] = 1. Let [0, 1] denote the closed unit interval with the From Proposition A4.1.11, it suﬃces to show that 0 = H1[0, 1] < ∞. Proof. This is the case if we show H1[0, 1] = 1. For any 1 > ε > 0, it is clear that the interval [0, 1] can be covered by nε ε[0, 1] ε(2 + 1/ε); ε[0, 1] 1 + 2ε. Thus H1[0, 1] 1. From Example A4.1.23, we now have intervals each of diameter less than ε, where nε 2 + 1/ε. So H1 that is, H1 H1[0, 1] = 1, from which the Proposition follows. A similar argument to that above shows that if a, b ∈ R with a < b, where R has the euclidean metric, then dimH[a, b] = 1. The next Corollary includes this result and is an easy consequence of combining Proposition A4.1.24, Example A4.1.18, Proposition A4.1.14, Proposition 4.3.5, and the deﬁnition of totally disconnected in Exercises 5.2 #10. A4.1.25 Corollary. Let R denote the set of all real numbers with the euclidean metric. (i) dimH R = 1. (ii) If S ⊂ R, then dimH S 1. (iii) If S contains a non-trivial interval (that is, is not totally disconnected), then dimH S = 1. (iv) If S is a non-trivial interval in R, then dimH S = 1. Proof. Exercise In fact if Rn has the euclidean metric, with n ∈ N, then it is A4.1.26 Remark. true that dimH Rn = n. This is proved in Exercises A4.1 #15. However, the proof there depends on the Generalized Heine-Borel Theorem 8.3.3 which is not proved until Chapter 8. This appendix on Hausdorﬀ dimension is not yet complete. From time to time please check for updates. 479 Exercises A4.1 1. Let Y be a subset of a metric space (X, d) and Y its closure. Prove that diam Y = diam Y . 2. Prove Proposition A4.1.14. [Hint. Use Deﬁnitions A4.1.4 and A4.1.10.] 3. Prove Lemma A4.1.15. n i=1 4. If Y = Yi, for some n ∈ N, is a separable subset of a metric space (X, d), show that dimH(Y ) = sup{dimH(Yi) : i = 1, 2 . . . , n}. 5. (i) Let n ∈ N, and a, b ∈ Rn. Show that if r and s are any positive real numbers, then the open balls Br(a) and Bs(b) in Rn with the euclidean metric satisfy dimH Br(a) = dimH Bs(b). (ii) Using the method of Example A4.1.18, show that dimH Br(a) = dim Rn. (ii) If S1 is the open cube {x1, x2, . . . , xn ∈ Rn :
0 < xi < 1, i = 1, . . . , n}, prove that dimH S1 = dimH Rn. (iii)* Using the method of Proposition A4.1.24, show that if n = 2 then H2(S1) 2 and so dimH(S1) 2. (iv) Prove that dimH R2 2. (v)* Using an analogous argument, prove that dimH Rn n, for all n > 2. 6. Prove Proposition A4.1.19. [Hint. Prove that as.Hs(X) Hs(Y ) bs.Hs(X).] 480 APPENDIX 4: HAUSDORFF DIMENSION 7. Let f : R → R2 be the function given by f (x) = x, x2. Using Proposition A4.1.19, show that dimH f [0, 1] = dimH[0, 1]. Deduce from this and Proposition A4.1.16 that if Y is the graph in R2 of the function θ : R → R given by θ(x) = x2, then dimH(Y ) = dimH[0, 1]. 8. Using an analogous argument to that in Exercise 7 above, show that if Z is the graph in R2 of any polynomial φ(x) = anxn + an−1xn−1 + . . . a2x2 + a1x + a0, where an = 0, then dimH Z = dimH[0, 1]. 9.* Let g : R → R be a function such that the nth-derivative g(n) exists, for each n ∈ N. Further assume that there exists a K ∈ N, \|g(n)(x)\| < K, for all n ∈ N and all x ∈ [0, 1]. (Examples of such functions include g = exp, g = sin, g = cos, and g is a polynomial.) Using the Taylor series expansion of g, extend the method of Exercises 7 and 8 above to show that if f : R → R2 is given by f (x) = x, g(x), then dimH f [0, 1] = dimH[0, 1]. 10. Prove Proposition A4.1.21. [Hint. Firstly prove that if z is any positive real number greater than 1, and Ui is any set in (X, d) of diameter less than ε, then there exists an open set Oi such that (i) Ui ⊆ Oi, (ii) diam Oi < ε, and (iii) diam Oi z. diam Ui. Use this to show that Os ε(Y ), for all z > 1.] ε(Y ) zs.Hs 11. Prove Lemma A4.1.22. [Hint. First assume that Y is covered by 2 open sets and prove the analogous result. Then consider the case that Y is covered by a ﬁnite number of open sets. Finally consider the case of an inﬁnite covering remembering a sum of an inﬁnite number of terms exists (and is ﬁnite) if and only if the limit of the ﬁnite sums exist.] 12. Show that if P denotes the set of all irrational numbers with the euclidean metric, then dimH P = 1 13. Fill in the details of the proof of Corollary A4.1.25. 481 14. The Generalized Heine-Borel Theorem 8.3.3 proved in Chapter 8, implies that if {Oi : i ∈ N} is an ε-covering of the open cube S1 of Example 5 above, then there exists an N ∈ N, such that {O1, O2, . . . , ON } is also an ε-covering of S1. Using this, extend Proposition A4.1.21 to say that: For every positive real number ε, Hs ε(S1) = inf (diam Oi)s   N  i=1    where N ∈ N and O1, . . . , ON is an open ε covering of S1. Warning: Note that this Exercise depends on a result not proved until Chapter 8. 15. (i) Show that if O is a subset of R2 with the euclidean metric, and A is its area, then A π 4 .(diam O)2. (ii) Deduce from (i) that if O1, O2, . . . , ON is an ε-covering of S1 in R2 of Example 5 above, then N i=1 (diam Oi)2 4 π . (iii) Deduce from (ii) and Exercise 14 above that H2(S1) 4 π . (iv) Using (iii) and Exercise 5, prove that dimH(S1) = dimH R2 = 2. (v) Using an analogous method to that above, prove that dimH Rn = n, where Rn has the euclidean metric. (vi) Prove that if S is any subset of Rn with the euclidean metric, such that S contains a non-empty open ball in Rn, the dimH S = n. Warning: Note that (iii), (iv), (v), and (vi) of this Exercise depend on a result proved in Chapter 8. Appendix 5: Topological Groups: A Graduate Course Contents of Appendix 5: Topological Groups §A5.0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 §A5.1 Topological Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 §A5.2 Subgroups and Quotient Groups of Topological Groups . . . . . . . . . . . . . . . 519 §A5.3 Embedding in Divisible Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 §A5.4 Baire Category and Open Mapping Theorems . . . . . . . . . . . . . . . . . . . . . . . . 534 §A5.5 Subgroups and Quotient Groups of Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 §A5.6 Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 §A5.7 Dual Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556 §A5.8 Pontryagin–van Kampen Duality Theorem: Introduction . . . . . . . . . . . . . . 562 §A5.9 Dual Groups of Subgroups, Quotients, and Finite Products . . . . . . . . . . . 565 §A5.10 Peter-Weyl Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 §A5.11 The Duality Theorem for Compact Groups and Discrete Groups . . . . . . 574 §A5.12 Monothetic LCA-groups and Compactly Generated LCA-groups . . . . . . . 580 §A5.13 The Duality Theorem and the Principal Structure Theorem . . . . . . . . . . . 590 §A5.14 Consequences of the Duality Theorem Part One: Kronecker’s Theorem 601 §A5.15 Consequences of the Duality Theorem Part Two: Connected and Totally Disconnected LCA-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607 §A5.99 Credit for Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ?? 482 483 484 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.0 Introduction In this Appendix we give an introduction to the theory of topological groups, the Pontryagin-van Kampen duality and structure theory of locally compact abelian groups11 It assumes that the reader is familiar with the notion of group as is included in an introductory course on group theory or usually in an introductory course on abstract algebra12. We shall begin by Meandering Through a Century of Topological Groups. This sets the stage for the study of topological groups, however the impatient reader can move onto §A5.1. 11Most of the material is this Appendix is taken from Morris [293]. 12A beautiful book on group theory is available as a free download. It is Macdonald [268]. In 1872 (Christian) Felix Klein, aged 23, was appointed to a full Professorship at Erlangen in Germany. In his ﬁrst year as Professor he published the ‘Erlangen Program13’. In his 1925 obituary of Klein, Richard Courant14 wrote, this is “perhaps the most inﬂuential and widely read paper in the second half of the nineteenth century”. In 1940 Julian Lowell Coolidge15 wrote that it “probably inﬂuenced geometrical thinking more than any other work since the time of Euclid, with the exception of Gauss and Riemann”. Klein was inﬂuenced by Julius Plücker (1801– 1868) and Rudolf Friedrich Alfred Clebsch(1833– 1872), both of whom had made signiﬁcant contributions to geometry. Klein completed his PhD in 1868 at the age of 19, shortly after his (ﬁrst) PhD adviser, Julius Plücker16, died. After Plücker’s death, Klein came under the inﬂuence of Rudolf Clebsch17, who assisted Klein [Clebsch obtain the Professorship at Erlangen. died in 1872 before the publication of the Erlangen Program.] When Clebsch died, most of his PhD students moved to Erlangen and became students18 of Klein. 485 Klein Plücker Clebsch 13‘Erlanger Programm’ in German, 14“Felix Klein”, Jahresbericht der Deutschen Math. Vereiningung 34 (1925), 197–213. 15“A History of Geometrical Methods”, Oxford: Clarendon Press, 1940. 16Plücker was a mathematical descendant of Carl Friedrich Gauss. Klein was also a mathematical descendant of Leonhard Euler via his second PhD adviser Rudolf Otto Sigismund Lipschitz 17Clebsch and Carl Neumann founded ‘Mathematische Annalen’ and Klein made it one of the best mathematics journals. 18Klein had a total of 57 PhD students and now has over 30,000 mathematical descendants [including Karl Hofmann with whom I have coauthored books on compact groups and pro-Lie groups]. 486 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE The Mathematical Geneaology Project www.genealogy.math.ndsu.nodak.edu has the following information, where we see for example that Kein supervised the PhD research of Lindemann who in turn supervised the PhD research of Hilbert. d’Alembert Euler Gauss Gerling Laplace Lagrange Poisson Dirichlet Fourier Plücker Klein Lipschitz Lindemann Hilbert Amongst these students who transferred to Klein was Ferdinand Lindemann, who in 1882 proved that π is a transcendental number; that is, π is not a root of any polynomial with rational number coeﬃcients. It was known that if π were transcendental, then the 2,000 year old problem of squaring the circle by compass and straightedge would be solved in the negative. Three ancient problems, (i) Delian problem – doubling the cube, i.e. constructing a cube of volume double that of a given cube, (ii) trisecting any given angle, (iii) squaring the circle (or quadrature of the circle), that is constructing a square whose area equals that of a given circle, in each case using only a straight edge and compass , fascinated professional & amateur mathematicians alike. (See Jones et al. [221].) 487 In 1775 the Paris Academy found it necessary to pass a resolution that no more solutions of any of these problems or of machines exhibiting perpetual motion were to be examined. Why did it take so long for these problems to be solved? Wantzel It was because an understanding came not from geometry but from abstract algebra (actually ﬁeld theory) a subject not born until the 19th century. The ﬁrst two problems were solved by Pierre Laurent Wantzel (1814–1848) in 1837 and the third by Carl Louis Ferdinand von Lindemann. Lindemann The ﬁeld theory used in solving these ancient problems was also used by Robert Henry Risch in the Risch Algorithm, https://en.wikipedia.org/wiki/Risch_algorithm, in 1968 which transforms the problem of indeﬁnite integration into a problem in algebra. The algorithm is partly implemented in the popular computer algebra package Maple19. So we see abstract algebra played a vital role in solving geometry
problems dating back 2,000 years and played a key role in the progress of computer algebra. Now we set the stage for the Erlangen Program of Felix Klein. We have all met Euclidean geometry which has points, lines, angles, and a metric (distance) and of course the famous Pythagoras theorem for right angled triangles which says that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides. In Euclidean geometry every two points determine a line However two lines do not necessarily determine a point: if they are not parallel, they do determine a point (of intersection) but if they are parallel, they don’t determine a point. This is a little disappointing. 19See www.maplesoft.com. 488 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE But a greater disappointment comes from art (or rather perspective : if we have a two dimensional picture/sketch/representation of a 3- dimensional object, then distances, angles, & parallel lines are not preserved. This can be easily seen in the two pictures on the next page. (See Credit for Images.) Brunelleschi In 1413 the Renaissance architect Filippo Brunelleschi introduced the geometrical method of perspective. In 1435, italian author, artist, architect, poet, priest, linguist, philosopher and cryptographer Leon Battista Alberti wrote Della pittura, a treatise on how to represent distance in painting. He used classical optics to explain perspective in art. Alberti 489 490 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE The branch of geometry dealing with the properties of geometric ﬁgures that remain invariant under projection is called projective geometry, and in earlier literature – descriptive geometry. Let V be a vector space. The projective space, A5.0.1 Deﬁnitions. P (V ), is the set of one-dimensional vector subspaces of V . If the vector space V has dimension n + 1, then P (V ) is said to be a projective space of dimension n. A one-dimensional projective space is called a projective line, and a two- dimensional projective space is called a projective plane. (See Examples 11.3.4 for a further discussion of the real projective plane and real projective space.) If V is the 3-dimensional vector space over the ﬁeld R of real numbers, the projective space P 2(R) has as its “points” the 1-dimensional vector subspaces and its “lines” are the 2-dimensional vector subspaces. A “point” S is said to lie on a “line” L if the space S is contained in the space L. Note that (i) any two distinct “points” determine a unique “line”; (ii) any two distinct “lines” determine a unique “point”. • There are no parallel “lines". • There is a duality between “points” & “lines”; theorems remain true with “point” and “line” interchanged. 491 u w , v w Using so-called homogeneous co-ordinates, we can represent points in the projective plane as the set of all triples (u, v, w), except (0, 0, 0), where u, v, w ∈ R, and for any c ∈ R \ {0}, the points (u, v, w) and (cu, cv, cw) are identiﬁed. Of these points (u, v, w), those with w = 0 can be regarded as points in the Euclidean plane, whereas the points (u, v, 0), can be thought of as points at inﬁnity. So the projective plane can be thought of as consisting of the euclidean plane plus points at inﬁnity. Formally this says, P 2(R) = R2 ∪ P 1(R). We are now in a position to say a few words about the Erlangen Program20 of Felix Klein. In 1869, Klein went to Berlin and in his own words “The most important event of my stay in Berlin was certainly that, toward the end of October, at a meeting of the Berlin Mathematical Society, I made the acquaintance of the Norwegian, Sophus Lie. Our work had led us from diﬀerent points of view ﬁnally to the same questions, or at least to kindred ones. Thus it came about that we met every day and kept up an animated exchange of ideas.” “The E.P. itself.. was composed in October, 1872. . . Lie visited me for two months beginning September 1. Lie, who on October 1 accompanied me to Erlangen. . . had daily discussions with me . . . entered eagerly into my idea of classifying the diﬀerent approaches to geometry on a grouptheoretic basis.”21 Erlangen Program 20See http://tinyurl.com/jxbctma for the original Erlangen Program and http://tinyurl.com/hb8pmtn for an English translation, and Klein [240] 21See Garrett Birkhoﬀ and M. K. Bennett, Felix Klein and his “Erlanger Programm", in History and Philosophy of Modern Mathematics, eds. W. Aspray and P. Kitcher, Minnesota Stud. Philos. Sci. XI, University of Minnesota Press, Minneapolis (1988), 145-176. http://tinyurl.com/8mjstzd 492 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Klein suggested that not only euclidean geometry, but also the “new” nonEuclidean geometries, can be regarded as “sub-geometries” of projective geometry22 He also suggested each geometry can be characterized by the invariants of its associated transformation group. In euclidean geometry, ﬁgures F1 & F2 are “equal” if they are congruent; i.e., F2 is obtained from F1 by a rigid motion (translation, rotation, & reﬂection) of euclidean space. So the transformation group associated with euclidean geometry is the group consisting of the transformations corresponding to rigid motions. Each geometry is similarly characterized by its symmetries using the group of geometric transformations of a space each of which take any ﬁgure onto an “equal” ﬁgure in that geometry. The set of geometric transformations should have certain natural properties, such as (1) every ﬁgure F is ‘equal’ to itself; (2) if a ﬁgure F1 is ‘equal’ to F2, then F2 is ‘equal’ to F1; and (3) if F1 is ‘equal’ to F2 and F2 is ‘equal’ to F3, then F1 is ‘equal’ to F3. So we are led to insist that the set of transformations is in fact a group. 22Klein proposed to mathematical physicists that even a moderate cultivation of the projective purview might bring substantial beneﬁts. A major success was when Klein’s colleague, Hermann Minkowski (1864–1909), showed that the essence of Einstein’s Special Theory of Relativity is captured by the (spacetime) geometry of the Lorentz group. (See https://en.wikipedia.org/wiki/Lorentz_group for a discussion of the Lorentz group.) Nobel Laureate Eugene Paul Wigner (1902–1995) advocated extending the Erlangen Program to physics and demonstrated that symmetries expose the deepest secrets of physics. Quark theory, (See https://en.wikipedia.org/wiki/Quark for a description of quarks.) and even the Higgs boson particle (See https://en.wikipedia.org/wiki/Higgs_boson.) theory, are consequences of this approach. See Amir D. Aczel, “It’s a Boson! The Higgs as the Latest Oﬀspring of Math & Physics”, http://tinyurl.com/bqaonns. 493 A5.0.2 Deﬁnitions. all a, b ∈ G, a · b ∈ G, is said to be a group if A set, G, together with an operation · , such that for (i) (a · b) · c = a · (b · c), ∀a, b, c ∈ G; [associativity] (ii) there exists an element 1 in G, such that 1 · a = a · 1 = a, for every element a ∈ G; The (unique) element 1 ∈ G with this property is said to be the identity of the group. (iii) for each a in G, there exists an element b in G such that a · b = b · a = 1. The (unique) element b is called the inverse of a in G and is written a−1. A group G is said to be an abelian group if a · b = b · a, for all a, b ∈ G. A5.0.3 Examples. We list some important examples: (i) the group, R, of all real numbers with the operation of addition; [Note that the set of real numbers with the operation of multiplication is not a group as the element 0 has no inverse.] (ii) the group, Q, of all rational numbers with the operation of addition; We call Q a subgroup of R as Q ⊂ R and the group operation on Q is the restriction of the group operation on R. A subgroup N of a group G is called a normal subgroup if for all n ∈ N and g ∈ G, gng−1 ∈ N . (iii) the subgroup, Z, of R consisting of all integers; (iv) the group, T, consisting of all complex numbers of modulus 1 (i.e. the set of numbers e2πix, 0 x < 1) with the group operation being multiplication of complex numbers. Then T is called the circle group. All of the above groups are abelian groups. For a plethora of examples of nonabelian groups, we turn to groups of matrices. Recall that an n × n matrix M in said to be nonsingular (or invertible) if there is an n × n matrix M −1 such that M M −1 = M −1M = I, where I is the n × n identity matrix (with 1s on the diagonal and 0 elsewhere). 494 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.0.4 Deﬁnition. The multiplicative group of all nonsingular n × n matrices with complex number entries is called the general linear group over C and is denoted by GL(n, C). GL(n, C) and its subgroups are called matrix groups. As matrix multiplication is not commutative (i.e., M1.M2 does not necessarily equal M2.M1), each GL(n, C) is a nonabelian group. The subgroup of GL(n, C) consisting of all matrices M A5.0.5 Deﬁnition. such that M.M t = M t.M = I (i.e., M −1 = M t), where t denotes transpose, is called the orthogonal group and is denoted by O(n, C). If G and H are groups, then G × H is a group with A5.0.6 Deﬁnition. the group operation being coordinatewise multiplication; i.e., if g1, g2 ∈ G and h1, h2 ∈ H, then (g1, h1) · (g2, h2) = (g1 · g2, h1 · h2), where · denotes the group operation on G, H and G × H. The set G × H with this operation is called the product group. If I is any index set, the inﬁnite product of groups Gi : i ∈ I, i∈I analogously. Gi, is deﬁned A5.0.7 Deﬁnitions. Let G and H be groups and f : G → H a map (i) f is called a homomorphism if f (g1 · g2) = f (g1) · f (g2), for all g1, g2 ∈ G. (ii) If f is also surjective (i.e., f (G) = H), then H is said to be a quotient group of G, written H = G/N , where the kernel N = {g ∈ G : f (g) = 1}. (iii) If f is bijective (i.e. injective [f (x) = f (y) ⇒ x = y] and surjective), then f is called an isomorphism. 495 Let G be a group with identity element 1, N a normal subgroup of G and H a subgroup of G. The following statements are equivalent
: (i) G = N H and N ∩ H = {1}. (ii) Every element of G can be written uniquely as a product of an element of N and an element of H. If one (= both) of these is true, then G is called a semidirect product of N and H, written G = NH. Semidirect product is more general than product; if H is also a normal subgroup, then N H = N × H. Pertinent to our earlier discussion are (i) the projective group, PGL(n, R) which is deﬁned as the quotient group GL(n, R)/K, where the kernel, K, is {λIn : λ ∈ R} and In denotes the n × n identity matrix; and (ii) the euclidean group, E(2), of all rigid motions of the plane, R2, is a semidirect product of the abelian group R2 (which describes translations) and the orthogonal group O(2, R) (which describes rotations and reﬂections which ﬁx the origin (0,0)). Felix Klein focussed on “discontinuous” groups and Sophus Lie focussed on “continuous” groups. Let G be a group with a topology τ . Then G is said to A5.0.8 Deﬁnition. be a topological group if the maps G → G given by g → g−1 and G × G → G given by (g1, g2) → g1 · g2 are continuous. 496 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.0.9 Examples. G is a topological group. topology then G is a compact topological group. (i) Let G be any group with the discrete topology. Then In particular, if G is any ﬁnite group with the discrete (ii) R, T, Z, Q, Ra × Tb × Zc, for any non-negative integers a, b and c with the usual topologies and the usual group operations are topological groups. (iii) For each n ∈ N, the group GL(n, C) and all of its subgroups can be regarded as and so have subspace topologies and with these topologies they subsets of Cn2 are topological groups. (iv) With the above topology, each O(n, C) is a compact group, which is Hausdorﬀ23. Topological groups G and H are said to be A5.0.10 Deﬁnition. topologically isomorphic if there exists a mapping f : G → H such that f is a homeomorphism and an isomorphism. This is written G ∼= H. A5.0.11 Deﬁnition. A topological group is said to be a compact Lie group if it is topologically isomorphic to a closed subgroup of an orthogonal group, O(n), for some n ∈ N. More generally, a Lie group is a group which is also a diﬀerentiable manifold, with the property that the group operations are compatible with the smooth structure. [See http://tinyurl.com/jeo75r7 for a description of diﬀerentiable manifolds.] (Marius) Sophus Lie 23All topological groups from here on in this introductory section are assumed to be Hausdorﬀ. The Lorentz group, already mentioned in the context of Einstein’s Special Relativity, and the Heisenberg group24 which plays a key role in Quantum Mechanics25 are Lie groups26 497 Lorentz27 Einstein28 Heisenberg29 Recall that a topological group G is called locally euclidean if it has an open set containing 1 which is homeomorphic to an open set containing 0 in Rn, n ∈ N. Note that for topological groups (i) compact ⇒ locally compact; (ii) locally euclidean ⇒ locally compact; (iii) Rn, n ∈ N, are not compact; (iv) Ra ×Tb ×Zc, a,b,c nonnegative integers, and all Lie groups are locally euclidean. In 1976 Jean Alexandre Eugène Dieudonné30(1906–1992) quipped “Les groupes de Lie sont devenus le centre de mathématique. On ne peut rien faire de sèrieux sans eux.” (Lie groups have moved to the centre of mathematics. One cannot seriously undertake anything without them.) Here “Lie theory” meant the structure theory of Lie algebras and Lie groups, and in particular how the latter is based on the former. Dieudonné 24See https://en.wikipedia.org/wiki/Heisenberg_group for a description of the Heisenberg group 25See http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.419.4136&rep=rep1&type=pdf for a discussion of the Heisenberg group and Quantum Mechanics. 26For a fascinating article on the Higgs Boson, and unifying General Relativity & Quantum Mechanics and their relation to the Exceptional Simple Lie Group E8, see A.G. Lisi and J.O. Weatherall, “A Geometric Theory of Everysthing”, Scientiﬁc American 303(2010) 54–61. 27Hendrik Antoon Lorentz (1853–1928) 28Albert Einstein (1879–1955) 29Werner Heisenberg (1901–1976) 30Dieudonné was a member of the Bourbaki group. See http://tinyurl.com/nl9k58s. 498 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE At the International Congress of Mathematicians in 1900, David Hilbert (862–1943) presented 23 problems that set the course for much of the mathematical creativity of the 20th century. Hilbert’s ﬁfth problem asked whether (in later terminology) a locally euclidean topological group is in fact a Lie group. It required half a century of eﬀort on the part of several generations of eminent mathematicians until it was settled in the aﬃrmative. Hilbert The most inﬂuential book on the solution of Hilbert’s ﬁfth problem and the structure of locally compact groups was Montgomery and Zippin [288]. A recent presentation of this work, and winner of the 2015 Prose Award for Best Mathematics Book, is by the UCLA31 academic, Australian born Flinders University32 graduate and 2006 Fields Medalist, Terence Chi-Shen Tao [389]33. Tao points out striking applications: Gromov’s celebrated theorem on groups of polynomial growth, and to the classiﬁcation of ﬁnite approximate groups and to the geometry of manifolds. Tao 31http://www.ucla.edu/ 32http://www.flinders.edu.au/ 33Terry Tao is a mathematical descendant of Leonhard Euler For 5 years Per Enﬂo looked at extending Hilbert’s Fifth Prolem to nonlocally compact groups and he said this “turned out be be useful for solving famous problems in Functional Analysis” – he solved “the approximation problem’ and the “basis problem” in Enﬂo [129] and the “invariant space problem” in Enﬂo [128]. For problem 153 in the Scottish Problem Book, https://en.wikipedia.org/wiki/Scottish_Book, which was later recognized as being closely related to Stefan Banach’s “basis problem", Stanisław Mazur oﬀered the prize of a live goose. This problem was solved only in 1972 by Per Enﬂo, who was presented with the live goose in a ceremony that was broadcast throughout Poland. 499 Per Enﬂo and Mazur, goose, and Enﬂo 500 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE At the beginning of the translation of the Scottish Problem Book, http://tinyurl.com/jmrmuuh, Stanisław Marcin Ulam writes: “The enclosed collection of mathematical problems has its origin in a notebook which was started in Lwów, in Poland in 1935. If I remember correctly, it was S. Banach who suggested keeping track of some of the problems occupying the group of mathematicians there. The mathematical life was very intense in Lwów. Some of us met practically every day, informally in small groups, at all times of the day to discuss .problems of common interest, communicating to each other the latest work and results. Apart from the more oﬃcial meetings of the local sections of the Mathematical Society (which took place Saturday evenings, almost every week!), there were frequent informal discussions mostly held in one of the coﬀee houses located near the University building - one of them a coﬀee house named “Roma”, and the other “The Scottish Coﬀee House". This explains the name of the collection. A large notebook was purchased by Banach and deposited with the headwaiter of the Scottish Coﬀee House, who, upon demand, would bring it out of some secure hiding place, leave it at the table, and after the guests departed, return it to its secret location. Many of the problems date from years before 1935, They were discussed a great deal among the persons whose names are included in the text, and then gradually inscribed into the book in ink.” . . . Banach Mazur Ulam 501 “As most readers will realize, the city of Lwów, and with it the Scottish Book, was fated to have a very stormy history within a few years of the book’s inception. A few weeks after the outbreak of World War II, the city was occupied by the Russians. From items at the end of this collection, it will be seen that some Russian mathematicians must have visited the town; they left several problems (and prizes for their solutions). The last date ﬁguring in the book is May Steinhaus 31, 1941. Item Number 193 contains a rather cryptic set of numerical results, signed by (Władysław Hugo Dionizy) Steinhaus, dealing with the distribution of the number of matches in a box! After the start of war between Germany and Russia, the city was occupied by German troops that same summer and the inscriptions ceased.” Partial solutions to Hilbert’s Fifth Problem came as the structure of topological groups was understood better: in 1923 Hermann Weyl (1885–1955) and his student Fritz Peter laid the foundations of the representation and structure theory of compact groups, and a positive answer to Hilbert’s Fifth Problem for compact groups was a Weyl consequence, drawn by John von Neumann (1903– Neumann.jpg von Neumann 1957) in 1932. [Peter-Weyl Theorem] Let G be any compact group. A5.0.12 Theorem. Then G is topologically isomorphic to a (closed) subgroup of a product of orthogonal groups. 502 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Lev Semyonovich Pontryagin (1908–1988) and Egbert Rudolf van Kampen (1908–1942) developed in 1932, resp., 1936, the duality theory of locally compact abelian groups laying the foundations for abstract harmonic analysis which ﬂourished throughout the second half of the 20th century. Pontryagin-van Kampen Duality34 provided the central method for attacking the structure theory of locally compact abelian groups. A positive response to Hilbert’s question for locally euclidean abelian groups followed in the wash. Pontryagin van Kampen If A is any abelian group, then the group Hom(A, T) A5.0.13 Deﬁnition. of all group homomorphisms of A into the circle group T (no continuity involved!) given the subspace topology from the product space TA (of all maps of A into T) is called the dual group of A and is written A. As the dual group is clearly a closed subset of TA, we obtain: A5.0.14 Proposition. The dual gr
oup of any abelian group is an abelian compact group. If G is any abelian compact group, then the abelian A5.0.15 Deﬁnition. group (without topology) Hom(G, T) of all continuous homomorphisms of G into T is called the dual group of the abelian compact group G and is written G. So if G is an abelian compact group, then its dual group, G, is an abelian group and the dual group of that dual group, G, is again an abelian compact group. 34See Morris [293]. Further there is a natural evaluation map from G into its second dual, namely η : G → G, where for each g ∈ G, η(g) = ηg : G → T and for each γ ∈ G, ηg(γ) = γ(g) ∈ T. 503 A5.0.16 Theorem. [Pontryagin-van Kampen Duality for Compact Groups] If G is any abelian compact group, then the evaluation map η : G → G is a topological group isomorphism. A5.0.17 Remark. The above theorem implies that no information about G is lost in going to the dual group. But the dual group is just an abelian group without topology. From this we deduce the fact that every piece of information about G can be expressed in terms of algebraic information about its dual group. So questions about compact abelian groups are reduced to ones about abelian groups. The next proposition provides some examples of this: A5.0.18 Proposition. Let G be an abelian compact group. (i) The weight, w(G), of G equals the cardinality of its dual group G; (ii) G is metrizable if and only if its dual group G is countable; (iii) G is connected if and only if its dual group G is torsion-free (i.e., it has no nontrivial ﬁnite subgroups); (iv) G is torsion-free if and only if its dual group G is divisible (i.e., if g ∈ ˆG, then there is an hn ∈ ˆG with (hn)n = g, for all positive integers n.) The next theorem is quite surprising. A5.0.19 Theorem. If abelian connected compact groups G1 and G2 are homeomorphic, then they are topologically isomorphic. A topological group G is said to be a protorus if it is compact connected and abelian. So Theorem A5.0.19 can be restated: Two protori are topologically isomorphic if and only if they are homeomorphic. 504 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE We now present the main structure theorem for locally compact abelian groups which follows from Pontryagin van-Kampen Duality and answers Hilbert’s ﬁfth problem for locally compact abelian groups. A5.0.20 Theorem. Every connected locally compact abelian group is topologically isomorphic to Rn × K, where K is a compact connected abelian group and n is a nonnegative integer. One of the most signiﬁcant papers on topological groups was published in 1949 by Kenkichi Iwasawa (1917–1998), 3 years before Hilbert’s Fifth Problem was ﬁnally settled by the concerted contributions of Andrew Mattei Gleason (1921–2008), Dean Montgomery (1909–1992), Leon Zippin (1905–1995), and Hidehiko Yamabe (1923–1960). Iwasawa Montgomery It was Iwasawa who recognized for the ﬁrst time that the structure theory of locally compact groups reduces to that of compact groups and Lie groups provided one knew that they happen to be approximated by Lie groups in the sense of projective limits, in other words, if they were pro-Lie groups. [See Hofmann and Morris [190].] And this is what Yamabe established in 1953 for all locally compact groups G which are almost connected (i.e., the quotient group of G by its connected identity component is compact), e.g. connected locally compact groups or compact groups. Zippin Gleason Yamabe In 1976 the Council of the Australian Mathematical Society resolved to bring a high calibre speaker from overseas for each Annual Meeting. The ﬁrst of these invited speakers, in 1977, was Karl Heinrich Hofmann35. 35Hofmann is a mathematical grandson of David Hilbert. Hofmann was aware that in 1957 Richard Kenneth Lashof (1922–2010) recognized that not only Lie groups, but any locally compact group G has a Lie algebra g. Hofmann believed that this observation was the nucleus of a complete and rich, although inﬁnite dimensional, Lie theory which had 505 Lashof never been exploited. In his 1977 AustMS invited lecture at La Trobe University, Hofmann made a bold proposal, and that was to extend Lie Theory from Lie groups to a much wider class of topological groups. The ﬁrst stage of this project was to extend the Lie Theory to all Compact Groups and this culminated in the publication of the ﬁrst edition of the 800 page book, Karl H. Hofmann Karl Hofmann & Sidney Morris at work and Sidney A. Morris, “The Structure of Compact Groups”, de Gruyter, 1998, with the 2nd edition in 2006, and the 3rd 900+ page edition, Hofmann and Morris [191], appearing in 2013. This book uses Lie Theory to expose the structure of compact groups, thereby proving old and new results by new methods. 506 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE In our book, Lie Theory is used to describe general compact groups in terms of their building blocks: (i) simple simply connected Lie groups (which are known. For a discussion, see http://tinyurl.com/hpqkjq2 in Wikipedia. The classiﬁcation of simple Lie groups was done by Wilhelm Killing and Élie Cartan.) Killing Cartan (ii) compact connected abelian groups (which are understood from Pontryagin-van Kampen Duality), and (iii) proﬁnite (⇔ compact totally disconnected) groups. For each topological group G deﬁne the topological space L(G) = Hom(R, G) of all continuous group homomorphisms from R to G, endowed with the topology of uniform convergence on compact sets. Deﬁne the continuous function exp : L(G) → G by exp X = X(1) and a “scalar multiplication” (r, X) → r.X : R × L(G) → L(G) by (r.X)(s) = X(sr). This is useful when G is such that L(G) is a Lie algebra with addition and In Hofmann and Morris [191], it is shown that bracket multiplication continuous. this occurs for all compact groups. 507 Having demonstrated conclusively the power of Lie Theory in the context of compact groups, the second stage was to apply it to locally compact groups. However, the category of locally compact groups is not a good one, as even inﬁnite products of locally compact groups are not locally compact except in the trivial case that they are almost all compact, e.g, even Rℵ0 is not locally compact. The category of all pro-Lie groups (and their continuous homomorphisms) Hofmann and Morris, August 2016 is well-behaved and includes all compact groups, all Lie groups, and all connected locally compact groups. Pro-Lie groups are deﬁned to be projective limits of Lie groups. Lie Theory was successfully applied to this class and the structure of these groups was fully exposed in the 600+ page book: Karl H. Hofmann and Sidney A. Morris, “The Lie Theory of Connected Pro-Lie Groups”, European Mathematical Society Publ. House, 2007, Hofmann and Morris [190]. Our next beautiful structure theorem describes the the structure of abelian connected pro-Lie groups completely and the topology of connected pro-Lie groups. Let G be a connected pro-Lie group. Then G is A5.0.21 Theorem. homeomorphic to RJ × C, for some set J and maximal (connected) compact subgroup, C, of G. If G is abelian, then G is topologically isomorphic to RJ ×C. The above theorem due to Karl H. Hofmann and Sidney A. Morris is a generalization of the classical result for connected locally compact groups. 508 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be a connected locally compact group. Then A5.0.22 Theorem. G is homeomorphic to Rn × C, for some set n ∈ N and maximal (connected) compact subgroup, C, of G. If G is abelian, then G is topologically isomorphic to Rn × C. We aim to describe the topology of compact groups in terms of simple simply connected compact Lie groups, compact connected abelian groups, and proﬁnite groups. Our next result is surprising. This theorem tells us everything about the topology of totally disconnected compact groups. A5.0.23 Theorem. The underlying topological space of every inﬁnite totally disconnected compact group is a Cantor cube. This result is not trivial, but it can be proved by elementary means or by an application of the structure theory of compact groups. We shall see that the above theorem does much more than describe the topology of a special class of compact groups. If G is any topological group, then the largest A5.0.24 Deﬁnition. connected set containing 1 is said to be the identity component of G and is denoted by G0. It is readily veriﬁed that for any topological group G, the identity component G0 is a closed normal subgroup of G. Further, if G is a compact group, then G0 is a compact group. The following proposition is obvious. Let G be a topological group. Then the quotient A5.0.25 Proposition. group G/G0 is a totally disconnected topological group. Further, if G is a compact group, then G/G0 is a totally disconnected compact group. Now we state a powerful result which signiﬁcantly reduces the task of describing the topology of a general compact group. 509 A5.0.26 Theorem. the product group G0 × G/G0. If G is any compact group then it is homeomorphic to Theorem A5.0.26 and Theorem A5.0.23 together reduce the study of the topology of compact groups to the study of the topology of connected compact groups. Let G be a topological group. Then G is said to A5.0.27 Deﬁnition. have no small subgroups or be an NSS-group if there exists an open set O containing the identity and O contains no non-trivial subgroup of G. For each positive integer n, the compact groups O(n), each discrete group, T, R, and Ra × Tb × Zc, for non-negative integers a, b, c are NSS-groups. A5.0.28 Theorem. [Hilbert 5 for Compact Groups] If G is a compact group, then the following conditions are equivalent: (i) G is a Lie group; (ii) G is an NSS-group; (iii) the topological space \|G\| is locally euclidean (that is, an open set containing 1 in G is homeomorphic to an open set containing 0 in Rn, for some positive integer n). Condition (iii) proves that compact Lie groups are characterized by just their topology. This is a beautiful result. The work of D. Montgomery,
L. Zippin and A. Gleason in the 1950s characterized noncompact Lie groups by conditions (ii) and (iii) above. Earlier we reduced the study of the topology of compact groups to the study of the topology of connected compact groups. 510 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Next we reduce the study to that of the topology of (a) abelian connected compact groups and what we shall call (b) semisimple groups. Let g, h be elements of a group G. Then g−1h−1gh ∈ A5.0.29 Deﬁnition. G is said to be a commutator and the smallest subgroup of G containing all commutators is called the commutator subgroup and denoted by G. A5.0.30 Theorem. If G is any connected compact group, then G is connected and (i) every element of G is a commutator, (ii) G is a compact group, and (iii) G = G. Condition (i) is remarkable; (ii) and (iii) are not valid without connectivity. A5.0.31 Deﬁnition. semisimple if G = G. A connected compact group G is said to be A5.0.32 Corollary. If G is any connected compact group, then G is semisimple. A5.0.33 Theorem. to G × G/G. If G is a connected compact group, it is homeomorphic 511 A5.0.34 Corollary. G/G0 × (G0) × G0/(G0), where G/G0 is homeomorphic to a Cantor cube. If G is any compact group then it is homeomorphic to We now state the Sandwich Theorem for Semisimple Connected Compact Groups. This tells us that each semisimple connected compact group is almost a product of simple simply connected Lie groups. Let G be a semisimple connected compact group. A5.0.35 Theorem. Then there is a family {Sj \| j ∈ J} of simple simply connected compact Lie groups and surjective continuous homomorphisms q and f j∈J Sj f −→ G q −→ Sj/Z(Sj) j∈J where each ﬁnite discrete abelian compact group Z(Sj) is the centre of Sj and qf −→ Sj j∈J j∈J Sj/Z(Sj) is the product of the quotient morphisms Sj → Sj/Z(Sj). A5.0.36 Remark. In conclusion, then, we have that every compact group is homeomorphic to the product of three groups whose topology we know: a compact totally disconnected group (which is homeomorphic to a Cantor cube), a compact connected abelian group, and a compact connected semisimple group. Using Theorem A5.0.21 we then have that every connected pro-Lie group is also homeomorphic to a product of three groups whose topology we know: RI for some index set I, a compact connected abelian group, and a compact connected semisimple group. This completes our meandering through a century of study of topological groups. Hopefully this overview of some of the highlights puts what follows in this appendix into context. The next section begins our formal study of topological groups. 512 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.1 Topological Groups Let (G, τ ) be a set G, that is a group, with a topology A5.1.1 Deﬁnition. τ on G. Then (G, τ ) is said to be a topological group if (i) the mapping (x, y) → xy of the product space (G, τ ) × (G, τ ) onto (G, τ ) is continuous, and (ii) the mapping x → x−1 of (G, τ ) onto (G, τ ) is continuous. 513 A5.1.2 Examples. (1) The additive group of real numbers with the euclidean topology is a topological group, usually denoted by R. (2) The multiplicative group of positive real numbers with the induced topology from R is also a topological group. (3) The additive group of rational numbers with the euclidean topology is a topological group denoted by Q. (4) The additive group of integers with the discrete topology is a topological group denoted by Z. (5) Any group with the discrete topology is a topological group. (6) Any group with the indiscrete topology is a topological group. (7) The “circle" group consisting of the complex numbers of modulus one (i.e. the set of numbers e2πix, 0 x < 1) with the group operation being multiplication of complex numbers and the topology induced from the euclidean topology on the complex plane is a topological group. This topological group is denoted by T (or S1). (8) Linear groups. Let A = (ajk) be an n × n matrix, where the coeﬃcients ajk are complex numbers. The transpose tA of the matrix A is the matrix (akj) and the conjugate A of A is the matrix (ajk), where ajk is the complex conjugate of the number ajk. The matrix A is said to be orthogonal if A = A and tA = A−1 and unitary if A−1 = t(A). The set of all non-singular n × n matrices (with complex number coeﬃcients) linear group (over the complex number ﬁeld) and is is called the general denoted by GL(n, C). The subgroup GL(n, C) consisting of those matrices with determinant one is the special linear group (over the complex ﬁeld) and is denoted by SL(n, C). The unitary group U (n) and the orthogonal group O(n) consist of all unitary matrices and all orthogonal matrices, respectively; they are subgroups of GL(n, C). Finally we deﬁne the special unitary group and the special orthogonal group as SU (n) = SL(n, C) ∩ U (n) and SO(n) = SL(n, C) ∩ O(n), respectively. The group GL(n, C) and all its subgroups can be regarded as subsets of Cn2 , where C denotes the complex number plane, and so Cn2 is a 2n2-dimension euclidean space. As such GL(n, C) and all its subgroups have induced topologies and it is easily veriﬁed that, with these, they are topological groups. 514 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.1.3 Remark. Of course not every topology on a group makes it into a topological group; i.e. the group structure and the topological structure need not be compatible. If a topology τ on a group G makes (G, τ ) into a topological group, then τ is said to be a group topology or a topological group topology. A5.1.4 Example. τ on G as follows: a subset U of G is open if either Let G be the additive group of integers. Deﬁne a topology (a) 0 ∈ U , or (b) G\U is ﬁnite. Clearly this is a (compact Hausdorﬀ ) topology, but Proposition A5.1.5 below shows that (G, τ ) is not a topological group. Let (G, τ ) be a topological group. For each a ∈ G, A5.1.5 Proposition. left and right translation by a are homeomorphisms of (G, τ ). Inversion is also a homeomorphism. Proof. The map La : (G, τ ) → (G, τ ) given by g → ag is the product of the two continuous maps (G, τ ) → (G, τ ) × (G, τ ) given by g → (a, g), where g ∈ G, a is ﬁxed, and (G, τ ) × (G, τ ) → (G, τ ) given by (x, y) → xy, x, y ∈ G, and is therefore continuous. So left translation by any a ∈ G is continuous. Further, = La has a continuous inverse, namely La−1, since La a(a−1g) = g and La−1 [La(g)] = La−1[ag] = a−1(ag) = g. So left translation is a homeomorphism. Similarly right translation is a homeomorphism. La−1(g) = La a−1g The map I : (G, τ ) → (G, τ ) given by g → g−1 is continuous, by deﬁnition. Also I has a continuous inverse, namely I itself, as I[I(g)] = I[g−1] = [g−1]−1 = g. So I is also a homeomorphism. It is now clear that the (G, τ ) in Example A5.1.4 above is not a topological group as left translation by 1 takes the open set {−1} onto {0}, but {0} is not an open set. What we are really saying is that any topological group is a homogeneous space while the example is not. Homogeneous spaces are deﬁned next. 515 A topological space (X, τ ) is said to be homogeneous A5.1.6 Deﬁnition. if it has the property that for each ordered pair x, y of points of X, there exists a homeomorphism f : (X, τ ) → (X, τ ) such that f (x) = y. While every topological group is a homogeneous topological space, we will see shortly that not ever homogeneous space can be made into a topological group. A5.1.7 Deﬁnition. A topological space is said to be a T1-space if each point in the space is a closed set. It is readily seen that any Hausdorﬀ space is a T1-space but that the converse is false. See Exercises 4.1 #13. We will see, however, that any topological group which is a T1-space is Hausdorﬀ. Incidentally, this is not true, in general, for homogeneous spaces–as any inﬁnite set with the coﬁnite topology is a homogeneous T1space but is not Hausdorﬀ. As a consequence we will then have that not every homogeneous space can be made into a topological group. A5.1.8 Proposition. identity element. neighbourhood V of e such that Let (G, τ ) be any topological group and e its If U is any neighbourhood of e, then there exists an open (i) V = V −1 (that is, V is a symmetric neighbourhood of the identity e) (ii) V 2 ⊆ U . (Here V −1 = {v−1 : v ∈ V } and V 2 = {v1v2 : v1 ∈ V, v2 ∈ V }, not the set {v2 : v ∈ V }.) Proof. Exercise. 516 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.1.9 Proposition. also a Hausdorﬀ space. Any topological group (G, τ ) which is a T1-space is Let x and y be distinct points of G. Then x−1y = e. The set G \ {x−1y} Proof. is an open neighbourhood of e and so, by Proposition A5.1.8, there exists an open symmetric neighbourhood V of e such that V 2 ⊆ G \ {x−1y}. Thus x−1y ∈ V 2. Now xV and yV are open neighbourhoods of x and y, respectively. Suppose xV ∩ yV = Ø. Then xv1 = yv2, where v1 and v2 are in V ; that is, x−1y = v1v−1 2 ∈ V.V −1 = V 2, which is a contradiction. Hence xV ∩ yV = Ø and so (G, τ ) is Hausdorﬀ. A5.1.10 Remark. So to check that a topological group is Hausdorﬀ it is only necessary to verify that each point is a closed set. Indeed, by Proposition A5.1.5, it suﬃces to show that {e} is a closed set. Warning. Many authors include “Hausdorﬀ" in their deﬁnition of topological group. A5.1.11 Remark. The vast majority of work on topological groups deals only with Hausdorﬀ topological groups. (Indeed many authors include “Hausdorﬀ" in their deﬁnition of topological group.) We will see one reason for this shortly. However, it is natural to ask: Does every group admit a Hausdorﬀ topology which makes it into a topological group? The answer is obviously “yes"–the discrete topology. But we mention the following problem. Question. Does every group admit a Hausdorﬀ non-discrete group topology which makes it into a topological group? Shelah [359] provided a negative answer, under the assumption of the continuum hypothesis. However in the special case that the group is abelian (that is, commutative) the answer is “yes" and
we shall prove this soon. Exercises A5.1 1. Let (G, τ ) be a topological group, e its identity element, and k any element of G. If U is any neighbourhood of e, show that there exists an open neighbourhood V of e such that 517 (i) V = V −1, (ii) V 2 ⊆ U , and (iii) kV k−1 ⊆ U . (In fact, with more eﬀort you can show that if K is any compact subset of (G, τ ) then V can be chosen also to have the property: (iv) for any k ∈ K, kV k−1 ⊆ U .) 2. (i) Let G be any group and let N = {N } be a family of normal subgroups of G. Show that the family of all sets of the form gN , as g runs through G and N runs through N is an open subbasis for a topological group topology on G. Such a topology is called a subgroup topology. (ii) Prove that every topological group topology on a ﬁnite group is a subgroup topology with N consisting of precisely one normal subgroup. 3. Show that (ii) if (G, τ ) is a topological group, then (G, τ ) is a regular space; (iii) any regular T0-space is Hausdorﬀ, and hence any topological group which is a T0-space is Hausdorﬀ. 4. Let (G, τ ) be a topological group, A and B subsets of G and g any element of G. Show that (i) If A is open, then gA is open. (ii) If A is open and B is arbitrary, then AB is open. (iii) If A and B are compact, then AB is compact. (iv) If A is compact and B is closed, then AB is closed. (v)* If A and B are closed, then AB need not be closed. 518 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 5. Let S be a compact subset of a metrizable topological group G, such that xy ∈ S if x and y are in S. Show that for each x ∈ S, xS = S. (Let y be a cluster point of the sequence x, x2, x3, . . . in S and show that yS = ∞ n=1 xnS; deduce that yxS = yS.) Hence show that S is a subgroup of G. (Cf. Hewitt and Ross [180], Theorem 9.16.) 6. A topological group G is said to be ω-narrow if for any neighbourhood N of the identity e, there exist a countable set xn ∈ G, n ∈ N, of members of G such that G = n∈N N xn. Verify each of the following statements: (i) Every countable topological group is ω-narrow. (ii) For each n ∈ N, Rn is ω-narrow. (iii) An uncountable discrete topological group is not ω-narrow. (iv) Every topological group which is a Lindelöf space is ω-narrow. In particular every topological group which is compact or a kω-space is ω-narrow. (v) Every topological group which is a second countable space is ω-narrow. (vi)* Every separable topological group is ω-narrow. [Hint. Let U be any open neighbourhood of e. Find a neighbourhood V of e such that V = V −1 and V 2 ⊆ U . Call a subset S of G V -disjoint if xV ∩ yV = Ø, for each distinct pair x, y ∈ S. The set S of all V -disjoint subsets of G is partially-ordered by set inclusion ⊆. As the union of any totally-ordered set of V -disjoint sets is a V -disjoint set, use Zorn’s Lemma to show there is a maximal element M of the partially-ordered set S. Verify that {mV : m ∈ M } is a disjoint set of non-empty open sets in G. Verify that in a separable space there is at most a countable number of disjoint open sets and hence M is countable. As M is maximal, show that for every x ∈ G, there exists an m ∈ M such that xV ∩ mV = Ø. Then x ∈ mV V −1 = mV 2 ⊆ mU. Deduce that M U = G and so G is ω-narrow.] A5.2 Subgroups and Quotient Groups of Topological Groups 519 Let G1 and G2 be topological groups. A map A5.2.1 Deﬁnition. f : G1 → G2 is said to be a continuous homomorphism if it is both a homomorphism of groups and continuous. If f is also a homeomorphism then it is said to be a topological group isomorphism or a topological isomorphism and G1 and G2 are said to be topologically isomorphic. Let R be the additive group of real numbers with the usual A5.2.2 Example. topology and R× the multiplicative group of positive real numbers with the usual topology. Then R and R× are topologically isomorphic, where the topological (Hence we need not mention this group R× isomorphism R → R× is x → ex. again, since, as topological groups, R and R× are the same.) A5.2.3 Proposition. G. With its subspace topology, H is a topological group. Let G be a topological group and H a subgroup of Proof. The mapping (x, y) → xy of H × H onto H and the mapping x → x−1 of H onto H are continuous since they are restrictions of the corresponding mappings of G × G and G. A5.2.4 Examples. (i) Z R; (ii) Q R. A5.2.5 Proposition. Let H be a subgroup of a topological group G. Then (i) the closure H of H is a subgroup of G; (ii) if H is a normal subgroup of G, then H is a normal subgroup of G; (iii) if G is Hausdorﬀ and H is abelian, then H is abelian. Proof. Exercise 520 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.2.6 Corollary. Let G be a topological group. Then (i) {e} is a closed normal subgroup of G; indeed, it is the smallest closed subgroup of G; (ii) if g ∈ G, then {g} is the coset g{e} = {e}g; (iii) If G is Hausdorﬀ then {e} = {e}. Proof. is a normal subgroup of G. This follows immediately from Proposition A5.2.5 (ii) by noting that {e} A5.2.7 Proposition. Any open subgroup H of a topological group G is (also) closed. Let xi, i ∈ I be a set of right coset representatives of H in G. So i∈I Hxi, where Hxi ∩ Hxj = Ø, for any distinct i and j in the index set I. Proof. G = Since H is open, so is Hxi open, for each i ∈ I. Of course for some i0 ∈ I, Hxi0 = H, that is, we have G = H∪ I\{i0}. These two terms are disjoint and the second term, being the union of open sets, is open. So H is the complement (in G) of an open set, and is therefore closed in G. i∈J Hxi , where J = Note that the converse of Proposition A5.2.7 is false. For example, Z is a closed subgroup of R, but it is not an open subgroup of R. Let H be a subgroup of a Hausdorﬀ group G. If H A5.2.8 Proposition. is locally compact, then H is closed in G. In particular this is the case if H is discrete. 521 Let K be a compact neighbourhood in H of e. Then there exists a Proof. neighbourhood U in G of e such that U ∩ H = K. In particular, U ∩ H is closed in G. Let V be a neighbourhood in G of e such that V 2 ⊆ U . If x ∈ H, then as H is a group (Proposition A5.2.5), x−1 ∈ H. So there exists an element y ∈ V x−1 ∩ H. We will show that yx ∈ H. As y ∈ H, this will imply that x ∈ H and hence H is closed, as required. To show that yx ∈ H we verify that yx is a limit point of U ∩ H. As U ∩ H is closed this will imply that yx ∈ U ∩ H and so, in particular, yx ∈ H. Let O be an arbitrary neighbourhood of yx. Then y−1O is a neighbourhood of x, and so y−1O ∩ xV is a neighbourhood of x. As x ∈ H, there is an element h ∈ (y−1O ∩ xV ) ∩ H. So yh ∈ O. Also yh ∈ (V x−1(xV ) = V 2 ⊆ U , and yh ∈ H; that is, yh ∈ O ∩ (U ∩ H). As O is arbitrary, this says that yx is a limit point of U ∩ H, as required. A5.2.9 Proposition. topological group G. Then H = Let U be a symmetric neighbourhood of e in a U n is an open (and closed) subgroup ∞ n=1 of G. Clearly H is a subgroup of G. Proof. Let h ∈ H. Then h ∈ U n, for some n. So h ∈ hU ⊆ U n+1 ⊆ H; that is, H contains the neighbourhood hU of h. As h was an arbitrary element of H, H is open in G. Proposition A5.2.7. It is also closed in G, by 522 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.2.10 Corollary. topological group G. Then G = generated by any neighbourhood of e. ∞ n=1 Let U be any neighbourhood of e in a connected U n; that is, any connected group is Proof. Let V be a symmetric neighbourhood of e such that V ⊆ U . By Proposition A5.2.9, H = V n is an open and closed subgroup of G. ∞ n=1 As G is connected, H = G; that is G = ∞ n=1 As V ⊆ U , V n ⊆ U n, for each n and so G = V n. ∞ n=1 U n, as required. A topological group G is said to be compactly A5.2.11 Deﬁnition. generated if there exists a compact subset X of G such that G is the smallest subgroup (of G) containing X. A5.2.12 Examples. (i) R is compactly generated by [0, 1] (or any other non-trivial compact interval). (ii) Of course, any compact group is compactly generated. A5.2.13 Corollary. Any connected locally compact group is compactly generated. Proof. ∞ n=1 G = Let K be any compact neighbourhood of e. Then by Corollary A5.2.10, Kn; that is, G is compactly generated. A5.2.14 Remark. In due course we shall describe the structure of compactly generated locally compact Hausdorﬀ abelian groups. We now see that this class includes all connected locally compact Hausdorﬀ abelian groups. Notation. By LCA-group we shall mean locally compact Hausdorﬀ abelian topological group. 523 A5.2.15 Proposition. The component of the identity (that is, the largest connected subset containing e) of a topological group is a closed normal subgroup. Proof. any topological space components are closed sets, C is closed. Let C be the component of the identity in a topological group G. As in Let a ∈ C. Then a−1C ⊆ C, as a−1C is connected (being a homeomorphic image of C) and contains e. So a∈C a−1C = C−1C ⊆ C, which implies that C is a subgroup. To see that C is a normal subgroup, simply note that for each x in G, x−1Cx is a connected set containing e and so x−1Cx ⊆ C. Let N be a normal subgroup of a topological group A5.2.16 Proposition. G. If the quotient group G/N is given the quotient topology under the canonical homomorphism p : G → G/N (that is, U is open in G/N if and only if p−1(U ) is open in G), then G/N becomes a topological group. Further, the map p is not only continuous but also open. (A map is said to be open if the image of every open set is open.) The veriﬁcation that G/N with the quotient topology is a topological Proof. group is routine. That the map p is continuous is obvious (and true for all quotient maps of topological spaces). To see that p is an open map, let O be an open set in G. Then p−1(p(O)) = N O ⊆ G. Since O is open, N O is open. (See Exercises A5.2 #4.) So by the deﬁnition of the quotient topology on G/N , p(O) is open in G/N ; that is, p is an open map. 524 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.2.17 Remarks. (i) Note that quotient maps of topological spaces are not necessarily open maps. (ii) Q
uotient maps of topological groups are not necessarily closed maps. For if R2 denote the product group R × R with the usual topology, example, and p is the projection of R2 onto its ﬁrst factor R, then the set S = is closed in R2 and p is a quotient map with p(S) not x, 1 x closed in R. : x ∈ R, x = 0 If G is a topological group and N is a compact A5.2.18 Proposition. normal subgroup of G then the canonical homomorphism p : G → G/N is a closed map. The homomorphism p is also an open map. If S is a closed subset of G, then p−1(p(S)) = N S which is the product Proof. in G of a compact set and a closed set. By Exercises A5.1 #4 then, this product is a closed set. So p(S) is closed in G/N and p is a closed map. As p is a quotient mapping, Proposition A5.2.16 implies that it is an open map. A5.2.19 Deﬁnition. A topological space is said to be totally disconnected if the component of each point is the point itself. If G is any topological group and C is the A5.2.20 Proposition. component of the identity, then G/C is a totally disconnected topological group. Proof. Note that C is a normal subgroup of G and so G/C is a topological group. The proof that G/C is totally disconnected is left as an exercise. A5.2.21 Proposition. group G, then G/N is locally compact. If G/N is any quotient group of a locally compact 525 Proof. Simply observe that any open continuous image of a locally compact space is locally compact. Let G be a topological group and N a normal A5.2.22 Proposition. subgroup. Then G/N is discrete if and only if N is open. Also G/N is Hausdorﬀ if and only if N is closed. Proof. This is obvious (noting that a T1-group is Hausdorﬀ ). Exercises A5.2 1. Let G and H be topological groups and f : G → H a homomorphism. Show that f is continuous if and only if it is continuous at the identity; that is, if and only if for each neighbourhood U in H of e, there exists a neighbourhood V in G of e such that f (V ) ⊆ U . 2. Show that the circle group T is topologically isomorphic to the quotient group R/Z. 3. Let B1 and B2 be (real) Banach spaces. Verify that (i) B1 and B2, with the topologies determined by their norms, are topological groups. (ii) If T : B1 → B2 is a continuous homomorphism (of topological groups) then T is a continuous linear transformation. (So if B1 and B2 are “isomorphic as topological groups" then they are “isomorphic as topological vector spaces" but not necessarily “isomorphic as Banach spaces".) 526 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 4. Let H be a subgroup of a topological group G. Show that H is open in G if and only if H has non-empty interior (that is, if and only if H contains a non-empty open subset of G). 5. Let H be a subgroup of a topological group G. Show that (i) H is a subgroup of G. (ii) If H is a normal subgroup of G, then H is a normal subgroup of G. (iii) If G is Hausdorﬀ and H is abelian, then H is abelian. 6. Let Y be a dense subspace of a Hausdorﬀ space X. If Y is locally compact, show that Y is open in X. Hence show that a locally compact subgroup of a Hausdorﬀ group is closed. 7. Let C be the component of the identity in a topological group G. Show that G/C is a Hausdorﬀ totally disconnected topological group. Further show that if f is any continuous homomorphism of G into any totally disconnected topological group H, then there exists a continuous homomorphism g : G/C → H such that gp = f , where p is the projection p : G → G/C. 8. Show that the commutator subgroup [G, G] of a connected topological group G is connected. ([G, G] is generated by {g−1 1 g−1 2 g1g2 : g1, g2 ∈ G}.) 9. If H is a totally disconnected normal subgroup of a connected Hausdorﬀ group G, show that H lies in the centre, Z(G), of G (that is, gh = hg, for all g ∈ G and h ∈ H). [Hint: Fix h ∈ H and observe that the map g → ghg−1 takes G into H.] 10. (i) Let G be any topological group. Verify that G/{e} is a Hausdorﬀ topological 527 group. Show that if H is any Hausdorﬀ group and f : G → H is a continuous homomorphism, then there exists a continuous homomorphism g : G/{e} → H such that gp = f , where p is the canonical map p : G → G/{e}. (This result is the usual reason given for studying Hausdorﬀ topological groups rather than arbitrary topological groups. However, the following result which says in eﬀect that all of the topology of a topological group lies in its “Hausdorﬃzation", namely G/{e}, is perhaps a better reason.) (ii) Let Gi denote the group G with the indiscrete topology and i : G → Gi the identity map. Verify that the map p × i : G → G/{e} × Gi, given by p × i(g) = (p(g), i(g)), is a topological group isomorphism of G onto its image p × i(G). 11. Show that every Hausdorﬀ group, H, is topologically isomorphic to a closed subgroup of an arcwise connected, locally arcwise connected Hausdorﬀ group G. (Consider the set G of all functions f : [0, 1) → H such that there is a sequence 0 = a0 < a1 < a2 < · · · < an = 1 with f being constant on each [ak, ak−1). Deﬁne a group structure on G by f g(t) = f (t)g(t) and f −1(t) = (f (t))−1, where f and g ∈ G and t ∈ [0, 1). The identity of G is the function identically equal to e in H. For ε > 0 and any neighbourhood V of e in H let U (V, ε) be the set of all f such that λ({t ∈ [0, 1) : f (t) ∈ V }) < ε, where λ is Lebesgue measure on [0, 1). The set of all U (V, ε) is an open basis for a group topology on G. The constant functions form a closed subgroup of G topologically isomorphic to H.) 528 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 12. Verify the following statements. (i) Every subgroup of an ω-narrow topological group is ω-narrow. (ii) Let G and H be topological groups and f : G → H a continuous If G is an ω-narrow topological group, then H is an ω- homomorphism. narrow topological group. (iii) Let I be an index set and for each i ∈ I, let Gi be an ω-narrow topological group. Then the product i∈I Gi is an ω-narrow topological group. (iv) Exercises A5.1 #6(vi) says that every separable topological group is an ωnarrow topological group. However, it follows from (iii) above that there exist non-separable ω-narrow topological groups. (v) Let the topological group G be a subgroup of a (ﬁnite or inﬁnite) product of topological groups each of which is a second countable space. By Exercises A5.1 #6 and (i) and (iii) above, G is an ω-narrow topological group. [It is proved in Theorem 3.4.23 of Arhangel’skii and Tkachenko [15] that a topological group is ω-narrow if and only if it is topologically isomorphic to a subgroup of a product of second countable topological groups.] A class V of topological groups is said to be a variety of topological groups if (i) every subgroup of a member of V is a member of V (ii) every quotient group of a member of V is a member of V and (iii) every product of a set of members of V is a member of V. (See Morris [295].) So we see that the class of ω-narrow topological groups is a variety of topological groups. Other examples are the class of all topological groups and the class of all abelian topological groups. A5.3 Embedding in Divisible Groups 529 A5.3.1 Remark. Products of topological spaces are discussed in detail in Chapters 8, 9 and 10. The most important result on products is, of course, Tychonoﬀ ’s Theorem 10.3.4 which says that any (ﬁnite or inﬁnite) product (with the product topology) of compact topological spaces is compact. Further, Theorem 10.3.4 says that a product of topological spaces {(Xi, τ i) : i ∈ I} is compact only if each of the spaces (Xi, τ i) is compact. If each Gi is a group then i∈I (gihi), where gi and hi ∈ Gi). i∈I Gi has the obvious group structure ( i∈I gi · i∈I hi = i∈I If {Gi : i ∈ I} is a family of groups then the restricted direct product (weak rGi, is the subgroup of Gi consisting of elements i∈I direct product), denoted i∈I gi, with gi = e, for all but a ﬁnite number of i ∈ I. From now on, if {Gi : i ∈ I} is a family of topological groups then Gi will i∈I rGi will denote the denote the direct product with the product topology. Further i∈I restricted direct product with the topology induced as a subspace of i∈I Gi. A5.3.2 Proposition. If each Gi, i ∈ I is a topological group, then i∈I Gi is a topological group. Further i∈I rGi is a dense subgroup of i∈I Gi. Proof. Exercise. 530 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.3.3 Proposition. Let {Gi : i ∈ I} be a family of topological groups. Then Gi is locally compact if and only if each Gi is locally compact and all (i) i∈I but a ﬁnite number of Gi are compact. rGi is locally compact Hausdorﬀ if and only if each Gi is locally compact (ii) i∈I Hausdorﬀ and Gi = {e} for all but a ﬁnite number of Gi. Proof. Exercise. To prove the result we foreshadowed: every inﬁnite abelian group admits a non-discrete Hausdorﬀ group topology we need some basic group theory. A5.3.4 Deﬁnition. {xn : x ∈ D} = D; that is, every element of D has an nth root. A group D is said to be divisible if for each n ∈ N, A5.3.5 Examples. the group Z is not divisible. It is easily seen that the groups R and T are divisible, but If φ A5.3.6 Proposition. is any homomorphism of H into a divisible abelian group D, then φ can be extended to a homomorphism Φ of G into D. Let H be a subgroup of an abelian group G. By Zorn’s Lemma 10.2.16, it suﬃces to show that if x ∈ H, φ can be Proof. extended to the group H0 = {xnh : h ∈ H, n ∈ Z}. Case (i). Assume xn ∈ H, n ∈ N. Then deﬁne Φ(xnh) = φ(h), for all n ∈ Z. Clearly Φ is well-deﬁned, a homomorphism, and extends φ on H. Case (ii). Let k 2 be the least positive integer n such that xn ∈ H. So φ(xk) = d ∈ D. As D is divisible, there is a z ∈ D such that zk = d. Deﬁne Φ(xnh) = φ(h)zn, for all n ∈ Z. Clearly Φ is well-deﬁned, a homomorphism and extends φ on H. If G is an abelian group, then for any g and h in G, with A5.3.7 Corollary. g = h, there exists a homomorphism φ : G → T such that φ(g) = φ(h); that is, φ separates points of G. 531 Clearly it suﬃces to show that for each g = e in G, there exists a Proof. homomorp
hism φ : G → T such that φ(g) = e. Case (i). Assume gn = e, and gk = e for 0 < k < n. Let H = {gm : m ∈ Z}. Deﬁne φ : H → T by φ(g) = an nth root of unity = r, say, (r = e), and φ(gm) = rm, for each m. Now extend φ to G by Proposition A5.3.6. Case (ii). Assume gn = e, for all n > 0. Deﬁne φ(g) = z, for any z = e in T. Extend φ to H and then, by Proposition A5.3.6, to G. For later use we also record the following corollary of Proposition A5.3.6. Let H be an open divisible subgroup of an abelian A5.3.8 Proposition. topological group G. Then G is topologically isomorphic to H × G/H. (Clearly G/H is a discrete topological group.) Proof. Exercise. A5.3.9 Theorem. If G is any inﬁnite abelian group, then G admits a non-discrete Hausdorﬀ group topology. Let {φi : i ∈ I} be the family of distinct homomorphisms of G into T. Ti by putting Proof. Put H = i∈I f (g) = φi(g). Since each φi is a homomorphism, f is also a homomorphism. By i∈I Ti, where each Ti = T. Deﬁne a map f : G → H = i∈I Corollary A5.3.7, f is also one-one; that is, G is isomorphic to the subgroup f (G) of H. As H is a Hausdorﬀ topological group, f (G), with the topology induced from H, is also a Hausdorﬀ topological group. It only remains to show that f (G) is not discrete. 532 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Suppose f (G) is discrete. Then, by Proposition A5.2.8, f (G) would be a closed subgroup of H. But by Tychonoﬀ ’s Theorem 10.3.4, H is compact and so f (G)would be compact; that is, f (G) would be an inﬁnite discrete compact space– which is impossible. So we have a contradiction, and thus f (G) is not discrete. A5.3.10 Remark. Corollary A5.3.7 was essential to the proof of Theorem A5.3.9. This Corollary is a special case of a more general theorem which will be discussed later. We state the result below. If G is any LCA-group, then for any g and h in G, A5.3.11 Theorem. with g = h, there exists a continuous homomorphism φ : G → T such that φ(g) = φ(h). 1. If {Gi : i ∈ I} is a family of topological groups, show that Exercises A5.3 Gi is a topological group; rGi is a dense subgroup of i∈I (i) i∈I (ii) i∈I (iii) i∈I but a ﬁnite number of Gi are compact; Gi; Gi is locally compact if and only if each Gi is locally compact and all rGi is locally compact Hausdorﬀ if and only if each Gi is locally compact (iv) i∈I Hausdorﬀ and Gi = {e} for all but a ﬁnite number of Gi. 2. Show that if G is an abelian topological group with an open divisible subgroup H, then G is topologically isomorphic to H × G/H. 3. Let G be a torsion-free abelian group (that is, gn = e for each g = e in G, and each n ∈ N). Show that if g and h are in G with g = h, then there exists a homomorphism φ of G into R such that φ(g) = φ(h). 533 4. Let G be a locally compact totally disconnected topological group. (i) Show that there is a neighbourhood base of the identity consisting of compact open subgroups. (Hint: You may assume that any locally compact Hausdorﬀ totally disconnected topological space has a base for its topology consisting of compact open sets.) (ii) If G is compact, show that the “subgroups" in (i) can be chosen to be normal. (iii) Hence show that any compact totally disconnected topological group is topologically isomorphic to a closed subgroup of a product of ﬁnite discrete groups. (Hint: Let {Ai consisting of open normal subgroups. Let φi the canonical homomorphisms, and deﬁne Φ : G → i∈I Φ(g) = i∈I : i ∈ I} be a base of neighbourhoods of the identity : G → G/Ai, i ∈ I, be (G/Ai) by putting φi(gi).) 5. Let f : R → T be the canonical map and θ any irrational number. On the 1, x topological space G = R2 × T2 deﬁne an operation. 2, t1+t (x1, x2, t1, t2)·(x 1, x2+x Show that, with this operation, G is a topological group and that the commutator subgroup of G is not closed in G. (The commutator subgroup of a group G is the subgroup of G generated by the set {g−1h−1gh : g, h ∈ G}.) 2) = (x1+x 2+f (θx2x 1+f (x2x 1), t2+t 1, t 2, t 1)). 6. Let I be a set directed by a partial ordering . For each i ∈ I, let there be given a Hausdorﬀ topological group Gi. Assume that for each i and j in I such that i < j, there is an open continuous homomorphism fji of Gj into Gi. Assume further that if i < j < k, then fki = fjifkj. The object consisting of I, the groups Gi and the mappings fji, is called an inverse mapping system or a projective mapping system. The subset H of the product group G = Gi i∈I consisting of all i∈I (xi) such that if i < j then xi = fji(xj) is called the injective limit or projective limit of the inverse mapping system. Show that H is a closed subgroup of G. 534 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.4 Baire Category and Open Mapping Theorems A5.4.1 Theorem. (Baire Category Theorem for Locally Compact If X is a locally compact regular space, then X is not the union of a Spaces) countable collection of closed sets all having empty interior. ∞ n=1 Proof. Suppose that X = An, where each An is closed and Int(An) = φ, for that An. ∞ n=1 each n. Put Dn = X\An. Then each Dn is open and dense in X. We shall show Dn = Ø, contradicting the equality X = ∞ n=1 Let U0 be a non-empty open subset of X such that U0 is compact. As D1 is dense in X, U0 ∩ D1 is a non-empty open subset of X. Using the regularity of X we can choose a non-empty open set U1 such that U1 ⊆ U0 ∩ D1. Inductively deﬁne Un so that each Un is a non-empty open set and Un ⊆ Un−1 ∩ Dn. Since U0 is Dn = Ø. This compact and each Un is non-empty, ∞ n=1 contradiction the supposition is false and so the theorem is proved. Un = Ø. This implies ∞ n=1 A5.4.2 Remark. We saw that the Baire Category Theorem was proved for complete metric spaces in Theorem 6.5.1. The above Theorem also remains valid if “locally compact regular" is replaced by “locally compact Hausdorﬀ". A5.4.3 Corollary. topological group. Then G has the discrete topology. Let G be any countable locally compact Hausdorﬀ Proof. Exercise. A5.4.4 Theorem. (Open Mapping Theorem for Locally Compact Groups) Let G be a locally compact group which is σ-compact; that is, G = ∞ n=1 G onto a locally compact Hausdorﬀ group H. Then f is an open mapping. An, where each An is compact. Let f be any continuous homomorphism of 535 Let U be the family of all symmetric neighbourhoods of e in G and U Proof. the family of all neighbourhoods of e in H. It suﬃces to show that for every U ∈ U there is a U ∈ U such that U ⊆ f (U ). Let U ∈ U . Then there exists a V ∈ U having the property that V is compact and (V )−1V ⊆ U . The family of sets {xV : x ∈ G} is then an open cover of G and hence also of each compact set An. So a ﬁnite collection of these sets will cover any given An. So a ﬁnite collection of these sets will cover any given An. Thus there is a countable collection {xnV : n ∈ N} which covers G. ∞ n=1 f (xn)f (V ). This expresses H as a countable union of closed sets, and by the Baire Category Theorem A5.4.1, one of them must have non-empty interior; that is, f (xm)f (V ) contains an open set. Then f (V ) contains an open subset V of H. f (xnV ) = f (xnV ) = ∞ n=1 ∞ n=1 So H = To complete the proof select any point x of V and put U = (x)−1V . Then we have U = (x)−1V ⊆ (V )−1V ⊆ (f (V ))−1f (V ) = f ((V )−1V ) ⊆ f (U ), as required. A5.4.5 Remark. We met the Open Mapping Theorem for Banach Spaces in Theorem 6.5.5 Exercises A5.4 1. Show that any countable locally compact Hausdorﬀ group has the discrete topology. 536 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 2. Show that the Open Mapping Theorem A5.4.4 does not remain valid if either of the conditions “σ-compact” or “onto” is deleted. 3. Show that any continuous homomorphism of a compact group onto a Hausdorﬀ group is an open mapping. 4. Show that for any n ∈ N, the compact topological group Tn is topologically isomorphic to the quotient group Rn/Zn. 5. (i) Let φ be a homomorphism of a topological group G into a topological group H. If X is a non-empty subset of G such that the restriction φ : X → H is an open map, show that φ : G → H is also an open map. [Hint: For any subset U of G, φ(U ) = g∈G φ(U ∩ gX).] (ii) Hence show that if G and H are locally compact Hausdorﬀ groups with φ a continuous homomorphism : G → H such that for some compact subset K of G, φ(K) generates H algebraically, then φ is an open map. [Hint: Show that there is a compact neighbourhood U of e such that K ⊆ U . Put X = the subgroup generated algebraically by U .] 6. Let G and H be topological groups, and let η be a homomorphism of H into the group of automorphisms of G. Deﬁne a group structure on the set G × H by putting (g1, h1) · (g2, h2) = (g1η(h1)(g2), h1h2). Further, let (g, h) → η(h)(g) be a continuous map of G × H onto G. Show that (i) Each η(h) is a homeomorphism of G onto itself; and (ii) With the product topology and this group structure G × H is a topological group. (It is called the semidirect product of G by H that is determined by η, and is denoted by G η H.) 537 7. (i) Let G be a σ-compact locally compact Hausdorﬀ topological group with N a closed normal subgroup of G and H a closed subgroup of G such that G = N H and N ∩ H = {e}. Show that G is topologically isomorphic to an appropriately deﬁned semidirect product N η H. [Hint: Let η(h)(n) = h−1nh, h ∈ H and n ∈ N .] (ii) If H is also normal, show that G is topologically isomorphic to N × H. (iii) If A and B are closed compactly generated subgroups of a locally compact Hausdorﬀ abelian topological group G such that A ∩ B = {e} and G = AB, show that G is topologically isomorphic to A × B. 8. Let G and H be Hausdorﬀ topological groups and f a continuous homomorphism If G has a neighbourhood U of e such that U is compact and of G into H. f (U ) is a neighbourhood of e in H, show that f is an open map. A5.5 Subgroups and Quotient Groups of RRRn In this section we expose the structure of the closed subgroups and Hausdorﬀ quotient groups of Rn, n 1. Notation. Unless explicitly stated otherwise, for the remainder of this chapter we shall focus our attention on ab
elian groups which will in future be written additively. However, we shall still refer to the product of two groups A and B (and denote it by A × B) rather than the sum of the two groups. We shall also use An to denote the product of n copies of A and Ai for the product of the groups Ai, i ∈ I. i∈I The identity of an abelian group will be denoted by 0. A5.5.1 Proposition. Every non-discrete subgroup G of R is dense. Proof. We have to show that for each x ∈ R and each ε > 0, there exists an element g ∈ G ∩ [x − ε, x + ε]. As G is not discrete, 0 is not an isolated point. So there exists an element xε ∈ (G \ {0}) ∩ [0, ε]. Then the intervals [nxε, (n + 1)xε], n = 0, ±1, ±2, . . . cover R and are of length ε. So for some n, nxε ∈ [x−ε, x+ε] and of course nxε ∈ G. 538 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be a closed subgroup of R. Then G = {0}, A5.5.2 Proposition. G = R or G is a discrete group of the form aZ = {0, a, −a, 2a, −2a, . . . }, for some a > 0. Assume G = R. As G is closed, and hence not dense in R, G must be Proof. discrete. If G = {0}, then G contains some positive real number b. So [0, b] ∩ G is a closed non-empty subset of the compact set [0, b]. Thus [0, b] ∩ G is compact and discrete. Hence [0, b] ∩ G is ﬁnite, and so there exists a least element a > 0 in G. For each x ∈ G, let x a a < a. So x − x a denote the integer part of x a ∈ G a = 0; that is, x = na, for some n ∈ Z, as a . Then x − x a and 0 x − x a required. If a, b ∈ R then gp {a, b}, the subgroup of R generated A5.5.3 Corollary. by {a, b}, is closed if and only if a and b are rationally dependent. [Real numbers a and b are said to be rationally dependent if there exists integers n and m such that na = bm.] Proof. Exercise. A5.5.4 Examples. √ gp {1, √ 2} and gp { √ 2, 3} are dense in R. A5.5.5 Corollary. topologically isomorphic to T. Every proper Hausdorﬀ quotient group of R is If R/G is a proper Hausdorﬀ quotient group of R, then, by Proposition Proof. A5.2.22, G is a closed subgroup of R. By Proposition A5.5.2, G is of the form aZ, a > 0. Noting that the map x → 1 ax is a topological group isomorphism of R onto itself such that aZ maps to Z, we see that R/aZ is topologically isomorphic to R/Z which, we know, is topologically isomorphic to T. A5.5.6 Corollary. Every proper closed subgroup of T is ﬁnite. 539 Identify T with the quotient group R/Z and let p : R → R/Z be the Proof. canonical quotient homomorphism. If G is any proper closed subgroup of R/Z then p−1(G) is a proper closed subgroup of R. So p−1(G) is discrete. By Proposition A5.2.16, the restriction p : p−1(G) → G is an open map, so we see that G is discrete. As G is also compact, it is ﬁnite. We now proceed to the investigation of closed subgroups of Rn, for n 1. Here we use the fact that Rn is a vector space over the ﬁeld of real numbers. Notation. If A is a subset of Rn, we denote by sp R(A) the subgroup {α1a1 + · · · + αmam : αi ∈ R, ai ∈ A, i = 1, . . . , m, m a positive integer}; and by sp Q(A) the subgroup {α1a1 + · · · + αmam : αi ∈ Q, ai ∈ A, i = 1, . . . , m, m a positive integer}; and by gp (A) the subgroup of Rn generated by A. Clearly gp (A) ⊆ sp Q(A) ⊆ sp R(A). We deﬁne rank (A) to be the dimension of the vector space sp R(A). A5.5.7 Proposition. Rn, then gp {a1, . . . , am} is topologically isomorphic to Zm. If {a1, . . . , am} is a linearly independent subset of Choose elements am+1, . . . , an so that {a1, . . . , am, am+1, . . . , an} is a Proof. basis for Rn. It is clear that if {c1, . . . , cn} is the canonical basis for Rn, then gp {c1, . . . , cm} is topologically isomorphic to Zm. By Exercises A5.5#2, every linear transformation of Rn onto itself is a homeomorphism. So the linear map taking ai to ci, i = 1, . . . , n, yields a topological group isomorphism of gp {a1, . . . , am} onto gp {c1, . . . , cm} = Zm. 540 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.5.8 Proposition. p, and a1, . . . , ap ∈ G a basis for sp R(G). parallelotope with centre 0 and basis vectors a1, . . . , ap; P i=1 Let G be a discrete subgroup of Rn of rank Let P be the closed that is, P = . Then G∩P is ﬁnite and gp (G∩P ) = G. riai : −1 ri 1, i = 1, . . . , p Further, every point in G is a linear combination of {a1, . . . , ap} with rational coeﬃcients; that is, G ⊆ sp Q{a1, . . . , ap}. Proof. As P is compact and G is discrete (and closed in Rn), G ∩ P is discrete and compact, and hence ﬁnite. Now G ⊆ sp R{a1, . . . , ap} implies that each x ∈ G can be written as x = ti ∈ R. For each positive integer m, the point p i=1 tiai, zm = mx − p [mti]ai = p (mti − [mti])ai i=1 i=1 where [ ] denotes “integer part of", belongs to G. As 0 mti − [mti] < 1, zm ∈ P . Hence x = z1 + [ti]ai, which says that gp (G ∩ P ) = G. p i=1 Further, as G ∩ P is ﬁnite there exist integers h and k such that zh = zk. So (h − k)ti = [hti] − [kti], x ∈ sp Q{a1, . . . , ap}. A5.5.9 Corollary. Rn, and b = p i=1 t1, . . . , tp are rational numbers. Let {a1, . . . , ap} be a linearly independent subset of tiai, ti ∈ R. Then gp {a1, . . . , ap, b} is discrete if and only if Proof. Exercise. Every discrete subgroup G of Rn of rank p is generated A5.5.10 Theorem. by p linearly independent vectors, and hence is topologically isomorphic to Zp. 541 Since G is of rank p, G ⊆ sp R{a1, . . . , ap}, where a1, . . . , ap are linearly Proof. independent elements of G. By Proposition A5.5.8, G = gp {g1, . . . , gr} where each gi ∈ sp Q{a1, . . . , ap}. So there exists a d ∈ Z such that gi ∈ gp , i = 1, . . . , r. 1 da1, . . . , 1 dap Now, if {b1, . . . , bp} is a linearly independent subset of G, then bi = βijaj, where the determinant, det(βij) = 0, and βij ∈ 1 dp Z. So out of all d such {b1, . . . , bp} there exists one with det(βij) minimal. Let this set be denoted by {b1, . . . , bp}. We claim that G = gp {b1, . . . , bp} and hence is topologically isomorphic to Zp. Z. So det(βij) ∈ 1 Suppose G = gp {b1, . . . , bp}. Then there exists an element g ∈ G with p λibi and not all λi ∈ Z. Without loss of generality we can assume that g = i=1 λ1 = r s , r = 0 and s > 1. Since b1 ∈ G we can also assume that \|λ1\| < 1 (by subtracting multiples of b1, if necessary). Then putting b i = bi, i = 2, . . . , p and b 1 = g, b i = β ijaj we see that det(β ij) = det As \|λ1\| < 1 this means that           λ1 0 0 . . . 0 λ2 1 0 . . . 0 λ3 0 1 . . . 0 ... ... ... λp 0 0 . . . 1 < det(β ij) . . . ... det(βij) det(βij) = λ1 det(βij). , which is a contradiction. 542 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.5.11 Proposition. Every non-discrete closed subgroup H of Rn, n 1, contains a line through zero. As H is non-discrete there exists a sequence h1, h2, . . . of points in H Proof. converging to 0, with each hn = 0. Let C be an open cube with centre 0 containing all the hn. Let mn denote the largest integer m > 0 such that mhn ∈ C. The points mnhn, n = 1, 2, . . . lie in a compact set C and therefore have a cluster point a ∈ C ∩ H. If mnhn − a ε we have (mn + 1)hn − a ε + hn, where denotes the usual norm in Rn. Since hn → 0 as n → ∞ it follows that a is also a cluster point of the sequence (mn + 1)hn, n = 1, 2, . . . , whose points belong to the closed set Rn \ C. Hence a ∈ C ∩ (Rn \ C)–the boundary of C, which implies a = 0. Let t be any real number. Since \|tmn − [tmn]\| < 1, the relation mnhn − a ε implies that [tmn]hn − ta \|t\|ε + hn; since hn → 0 as n → ∞, ta is a limit point of the sequence [tmn]hn, n = 1, 2, . . . . But the points of this sequence belong to H and so ta ∈ H, since H is closed. So H contains the line through a = 0 and 0. A5.5.12 Theorem. are (closed) vector subspaces U , V and W of Rn such that Let G be a closed subgroup of Rn, n 1. Then there 543 (i) Rn = U × V × W (ii) G ∩ U = U (iii) G ∩ V is discrete (iv) G ∩ W = {0} (v) G = (G ∩ U ) × (G ∩ V ). Proof. that U is a vector subspace of Rn. Let U be the union of all lines through 0 lying entirely in G. We claim To see this let x and y be in U and λ, µ and δ ∈ R. Then δλx is in U and hence also in G. Similarly δµy ∈ G. So δ(λx + µy) = δλx + δµy ∈ G. As this is true for all δ ∈ R, we have that λx + µy ∈ U . So U is a vector subspace of Rn, and G ∩ U = U . Let U be any complementary subspace of U ; that is, Rn = U × U . So if g ∈ G, then g = h + k, h ∈ U , k ∈ U . As U ⊆ G, h ∈ G so k = g − h ∈ G. Hence G = U × (G ∩ U ). Put V = sp R(G ∩ U ) and W equal to a complementary subspace in U of V . So G ∩ W = {0}. Clearly G ∩ V contains no lines through 0, which by Proposition A5.5.11, implies that G ∩ V is discrete. If r A5.5.13 Theorem. equals the rank of G (that is, sp R(G) has dimension r) then there exists a basis a1, . . . , an of Rn such that Let G be a closed subgroup of Rn, n 1. G = sp R{a1, . . . , ap} × gp {ap+1, . . . , ar}. So G is topologically isomorphic to Rp × Zr−p and the quotient group Rn/G is topologically isomorphic to Tr−p × Rn−r. Proof. Exercise. 544 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Before stating the next theorem let us record some facts about free abelian groups. A5.5.14 Deﬁnition. A group F is said to be a free abelian group if it is the restricted direct product of a ﬁnite or inﬁnite number of inﬁnite cyclic groups. Each of these inﬁnite cyclic groups has a single generator and the set S of these generators is said to be a basis of F . A5.5.15 Remarks. (i) It can be shown that an abelian group F is a free abelian group with basis S if and only if S is a subset of F with the property that every map f of S into any abelian group G can be extended uniquely to a homomorphism of F into G. (ii) One consequence of (i) is that any abelian group G is a quotient group of some free abelian group. (Let F be the free abelian group with basis S of the same cardinality as G. Then there is a bijection φ of S onto G. Extend this map to a homomorphism of F onto G.) (iii) Proposition A5.5.7 together with Theorem A5.5.10 show that any subgroup of Zn is isomorphic to Zm, for some m. abelian group with ﬁnite basis is a fre
e abelian group with ﬁnite basis. In other words, any subgroup of a free It can be shown that any subgroup of a free abelian group is a free abelian group. For details see A.G. Kurosh [250]. (iv) Finally, we record that if the abelian group G admits a homomorphism φ onto a free abelian group F then G is isomorphic to F × A, where A is the kernel of φ. (Note that is suﬃces to produce a homomorphism θ of F into G such that φθ is the identity map of F . To produce θ, let S be a basis of F and for each s ∈ S choose a gs ∈ G such that φ(gs) = s. As F is a free abelian group the map s → gs of S into G can be extended to a homomorphism θ of F into G. Clearly φθ acts identically on F .) Let H = V × F , where V is a divisible abelian A5.5.16 Theorem. Hausdorﬀ group and F is a discrete free abelian group. If G is a closed subgroup of H, then there exists a discrete free abelian subgroup F of H isomorphic to F such that (i) H = V × F , and (ii) G = (G ∩ V ) × (G ∩ F ). 545 Let π1 : H → V and π2 : H → F be the projections. The restriction Proof. of π2 to G is a homomorphism from G to F with kernel G ∩ V . Since F is a free abelian group, and every subgroup of a free abelian group is a free abelian group, G/(G∩V ) is free abelian, and therefore, by the above Remark (iv), G is algebraically isomorphic to (G ∩ V ) × C, where C is a free abelian subgroup of G. Let p1 and p2 be the restrictions of π1 and π2 to C, respectively. Then p2 is one-one as 0}. F p2 ................................................................................................................. ............ ............ .............................................................................................. θ p1 C ................................................................................................................. ............ V We can deﬁne a homomorphism θ : p2(C) → V by putting θ(p2(c)) = p1(c) and then use Proposition A5.3.6 to extend θ to a homomorphism of F into the divisible group V . So θp2 = p1. If we now deﬁne a homomorphism φ : F → H by φ(x) = θ(x) + x and put F = φ(F ) we have that H = V × F , algebraically; the decomposition being given by v + f = [v − θ(f )] + [θ(f ) + f ], v ∈ V and f ∈ F. Also C ⊆ F , since for each c in C we have c = p1(c) + p2(c) = θ(p2(c)) + p2(c) = φ(p2(c)) ∈ φ(F ) = F . So (i) and (ii) are satisﬁed algebraically. Now φ : F → F is an algebraic isomorphism and since φ−1 is induced by π2, φ−1 is continuous. But F is discrete, so φ is a homeomorphism and F is a discrete free abelian group. 546 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE To show that H has the product topology with respect to the decomposition 1 : H → V 1(h) = 2(h) = π2(h) + θ(π2(h)), for each h ∈ H. Hence the H = V × F , it suﬃces to show that the corresponding projections π and π π1(h) − θ(π2(h)) and π decomposition G = (G ∩ V ) × (G ∩ F ) also has the product topology. 2 : H → F are continuous. But this is clearly the case since π Let G be a closed subgroup of Rn × Zm. Then G is A5.5.17 Corollary. topologically isomorphic to Ra × Zb, where a n and a + b n + m. Further, (Rn × Zm)/G is topologically isomorphic to Rc × Td × D, where D is a discrete ﬁnitely generated abelian group (with f m generators) and c + d n. Proof. Exercise. Let G be a closed subgroup of Rn × Tm × D, where D A5.5.18 Corollary. is a discrete abelian group. Then G is topologically isomorphic to Ra × Tb × D, where D is a discrete group and a + b n + m. Further (Rn × Tm × D)/G is topologically isomorphic to Rc × Td × D, where D is a discrete group and c + d n + m. Let F be a discrete free abelian group with D as a quotient group. (See Proof. Remarks A5.5.15.) Then there is a natural quotient homomorphism p of Rn+m × F onto Rn ×Tm ×D. So G is a quotient group of p−1(G) Rn+m ×F . Now Theorem A5.5.16 together with Theorem A5.5.13 describe both p−1(G) and the kernel of the map of p−1(G) onto G, and yield the result. In Corollary A5.5.18 we have not said that a n, b m A5.5.19 Remark. and c n. These inequalities are indeed true. They follow from the above and the Pontryagin-van Kempen Duality Theorem. Let G be a closed subgroup of Tn. Then G is A5.5.20 Corollary. topologically isomorphic to Ta×D where D is a ﬁnite discrete group and a n. 547 Proof. Exercise. The topological groups G and H are said to be locally A5.5.21 Deﬁnition. isomorphic if there are neighbourhoods V of e in G and U of e in H and a homeomorphism f of V onto U such that if x, y and xy all belong to V then f (xy) = f (x)f (y). A5.5.22 Example. R and T are obviously locally isomorpic topological groups. A5.5.23 Proposition. group G, then G and G/D are locally isomorphic. If D is a discrete normal subgroup of a topological Proof. Exercise. Let U be a neighbourhood of 0 in an abelian topological A5.5.24 Lemma. group G and V be a neighbourhood of 0 in Rn, n 1. If there is a continuous map f of V onto U such that x ∈ V , y ∈ V and x + y ∈ V implies f (x + y) = f (x) + f (y), then f can be extended to a continuous homomorphism of Rn onto the open subgroup of G generated by U . Proof. Exercise 548 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be a Hausdorﬀ abelian topological group locally A5.5.25 Theorem. isomorphic to Rn, n 1. Then G is topologically isomorphic to Ra × Tb × D, where D is a discrete group and a + b = n. By Lemma A5.5.24 there is a continuous homomorphism f of Rn onto Proof. an open subgroup H of G. As G is locally isomorphic to Rn, it has a compact neighbourhood of 0 and so is locally compact. Hence H is locally compact and the Open Mapping Theorem A5.4.4 says that f is an open map; that is, H is a quotient group of Rn. Further the kernel K of f is discrete since otherwise there would be elements x = 0 of K arbitrarily close to 0 such that f (x) = 0, which is false as f maps a neighbourhood of 0 homeomorphically into G. So Theorem A5.5.13 tells us that H is topologically isomorphic to Ra × Tb, with a + b = n. Now H is an open divisible subgroup of G which, by Proposition A5.3.8, implies that G is topologically isomorphic to H × D, where D = G/H is discrete. Thus G is topologically isomorphic to Ra × Tb × D, as required. The next corollary follows immediately. A5.5.26 Corollary. Rn, n 1, is topologically isomorphic to Ra × Tb, where a + b = n. Any connected topological group locally isomorphic to A5.5.27 Remark. We conclude this section by noting that some of the results presented here can be extended from ﬁnite to inﬁnite products of copies of R. For example, it is known that any closed subgroup of a countable product Ri of isomorphic copies Ri of R is topologically isomorphic to a countable product of isomorphic copies of R and Z. However, this result does not extend to uncountable products. For details of the countable products case, Brown et al. [60] and Leptin [258]. The uncountable case is best considered in the context of pro-Lie groups, Hofmann and Morris [190]. ∞ i=1 Exercises A5.5 549 1. If a, b ∈ R show that the subgroup of R generated by {a, b} is closed if and only if a and b are rationally dependent. 2. Prove that any linear transformation of the vector space Rn, n 1, onto itself is a homeomorphism. 3. (i) Let {a1, . . . , ap} be a linearly independent subset of Rn, n 1, and b = tiai, p i=1 ti ∈ R. Show that gp {a1, . . . , ap, b} is discrete if and only if t1, . . . , tp are rational numbers. (ii) Hence prove the following (diophantine approximation) result: Let θ1, . . . , θn be n real numbers. In order that for each ε > 0 there exist an integer q and n integers pi, i = 1, . . . , n such that \|qθi − pi\| ε, i = 1, . . . , n where the left hand side of at least one of these inequalities does not vanish, it is necessary and suﬃcient that at least one of the θi be irrational. 4. Prove Theorem A5.5.13 using the results preceding the Theorem. 5. Prove Corollary A5.5.17 using the results preceding the Corollary. 5. Prove Corollary A5.5.20 using the results preceding the Corollary. 6. Prove Proposition A5.5.23 using the results preceding the Proposition. 7. Prove that if a, b, n, m are integers with a + b = n + m and D1 and D2 are discrete groups, then Ra × Tb × D1 is locally isomorphic to Rn × Tm × D2. 8. Show that if G and H are locally isomorphic topological groups then there exists a neighbourhood V of e in G and U of e in H and a homeomorphism f of V onto U such that if x, y and xy all belong to V then f (xy) = f (x)f (y) and if x, y and xy all belong to U then f −1(xy) = f −1(x)f −1(y). 550 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 9. (i) Verify that any topological group locally isomorphic to a Hausdorﬀ topological group is Hausdorﬀ. (ii) Verify that any connected topological group locally isomorphic to an abelian group is abelian. (iii) Deduce that any connected topological group locally isomorphic to Rn, n 1, is topologically isomorphic to Ra × Tb, where a + b = n. 10. Prove Proposition A5.5.24 using the results preceding the Proposition. 11. Let U be a neighbourhood of 0 in an abelian topological group G and V a neighbourhood of 0 in Rn, n 1. If there is a continuous map f of V onto U such that x ∈ V , y ∈ V and x + y ∈ V implies f (x + y) = f (x) + f (y), show that f can be extended to a continuous homomorphism of Rn onto the open subgroup of G generated by U . A5.6 Uniform Spaces We now say a few words about uniform spaces just enough for our purposes here. For further discussion, see Kelley [233] and Bourbaki [51]. We introduce some notation convenient for this discussion. Let X be a set and X × X = X2 the product of X with itself. If V is a subset of X2 then V −1 denotes the set {(y, x) : (x, y) ∈ V } ⊆ X2. If U and V are subsets of X2 then U V denotes the set of all pairs (x, z), such that for some y ∈ X, (x, y) ∈ U and (y, z) ∈ V . Putting V = U deﬁnes U 2. The set {(x, x) : x ∈ X} is called the diagonal. A5.6.1 Deﬁnitions. subsets of X × X such that A uniformity on a set X is a non-empty set U of 551 (a) Each member of U contains the diagonal; (b
) U ∈ U =⇒ U −1 ∈ U ; (c) if U ∈ U then there is a V ∈ U such that V 2 ⊆ U ; (d) if U ∈ U and V ∈ U , then U ∩ V ∈ U ; (e) if U ∈ U and U ⊆ V ⊆ X2, then V ∈ U . The pair (X, U) is called a uniform space and each member of U is called an entourage. If R is the set of real numbers, then the usual uniformity A5.6.2 Examples. on R is the set U of all subsets of U of R × R such that {(x, y) : \|x − y\| < r} ⊆ U , for some positive real number r. Indeed if (X, d) is any metric space then we can deﬁne a uniformity U on X by putting U equal to the collection of all subsets U of X × X such that {(x, y) : d(x, y) < r} ⊆ U , for some positive real number r. Let (G, τ ) be a topological group and for each neighbourhood U of e, let UL = {(x, y) : x−1y ∈ U } and UR = {(x, y) : xy−1 ∈ U }. Then the left uniformity L on G consists of all sets V ⊆ G × G such that UL ⊆ V , for some U . Similarly we deﬁne the right uniformity. The two-sided uniformity consists of all sets W such that UL ⊆ W or UR ⊆ W , for some U . Given any uniformity U on a set X we can deﬁne a A5.6.3 Remarks. corresponding topology on X. For each x ∈ X, let Ux = {y ∈ X : (x, y) ∈ U }. Then as U runs over U , the system Ux deﬁnes a base of neighbourhoods at x for a topology; that is, a subset T of X is open in the topology if and only if for each x ∈ T there is a U ∈ U such that Ux ⊆ T . It is easily veriﬁed that if (G, τ ) is a topological group then the topologies arising from the left uniformity, the right uniformity and the two-sided uniformity all agree with the given topology. 552 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let (E, τ E) and (F, τ F ) be topological spaces and A5.6.4 Deﬁnitions. M any set of subsets M of E and {Vi : i ∈ I}, for some index set I, a base of open sets for the topology τ F on F . Let P (M, Vi) = {f : f ∈ F E and f (M ) ⊆ Vi}. (F E denotes the set of all functions f : E → F .) The family {P (M, Vi) : M ∈ M, I ∈ I}, is a subbase for a topology on F E . [If F is a Hausdorﬀ space and M is a covering of E then it is easily veriﬁed that this topology is Hausdorﬀ.] Two important special cases of this topology are when (a) M is the collection of all ﬁnite subsets of E – the topology is then the p-topology or the topology of pointwise convergence, and (b) M is the collection of all compact subsets of E – the k-topology or the compact-open topology. Since every ﬁnite set is compact, p ⊆ k. Therefore a subset of F E which is k-compact is also p-compact, but the converse if false. Observe that F E with the p-topology is simply x∈E topology, where each Fx is a homeomorphic copy of F . Fx, with the product We are interested in C(E, F ), the subset of F E consisting of all continuous functions from E to F , and we shall want to ﬁnd conditions which guarantee that a subset of C(E, F ) is k-compact. Let (E, τ E) and (F, τ F ) be topological spaces and G A5.6.5 Deﬁnition. a subset of F E . A topology τ on G is said to be jointly continuous if the map θ from the product space (G, τ ) × (E, τ E) to (F, τ F ), given by θ(g, x) = g(x), is continuous. A5.6.6 Proposition. continuous is ﬁner than the k-topology. Each topology τ on G ⊆ F E which is jointly 553 Let U be an open set in (F, τ F ), (K, τ K) a compact subspace of (E, τ E), Proof. and θ the map taking (g, x) to g(x), g ∈ G and x ∈ E. We want to show that for each f ∈ P (K, U ) = {g : g ∈ G and g(K) ⊆ U } there is a set W ∈ τ such that f ∈ W ⊆ P (K, U ). As θ is jointly continuous, the set V = (G × K) ∩ θ−1(U ) is open in (G, τ ), ×(K, τ K). If f ∈ P (K, U ), then {f } × K ⊆ V and since {f } × K is compact, there is a W ∈ τ such that f ∈ W and W × K ⊆ θ−1(U ). Hence W ⊆ P (K, U ) as required. A5.6.7 Proposition. G ⊆ C(E, F ). Then G is k-compact, that is compact in the k topology, if Let (E, τ E) and (F, τ F ) be topological spaces and (a) G is k-closed in C(E, F ), (b) the closure of the set {g(x) : g ∈ G} is compact in (F, τ F ), for each x ∈ E, and (c) the p-topology for the p-closure of G in F E is jointly continuous. Proof. Let G be the p-closure in F E of G. By condition (b), x∈E ({g(x) : g ∈ G}) is a p-compact set, and since G is a p-closed subset of this set, G is p-compact. By condition (c), the p-topology on G is jointly continuous – so G ⊆ C(E, F ). Also by Proposition A5.6.6, the p-topology on G is ﬁner than the k-topology and hence they coincide. Thus G is k-compact. As G is k-closed in C(E, F ) and G is k-compact and a subset of C(E, F ), we have that G is k-compact. 554 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let E be a topological space and F a uniform space. A5.6.8 Deﬁnitions. A subset G of C(E, F ) is said to be equicontinuous at the point x ∈ E if for each U in the uniformity U of F , there exists a neighbourhood V of x such that (g(y), g(x)) ∈ U , for all y ∈ V and g ∈ G. The family G is said to be equicontinuous if it is equicontinuous at every x ∈ E. A5.6.9 Proposition. at x ∈ E. Then the p-closure, G, in F E of G is also equicontinuous at x. Let G be a subset of C(E, F ) which is equicontinuous Proof. Exercise. A5.6.10 Proposition. Then the p-topology on G is jointly continuous. Let G be an equicontinuous subset of C(E, F ). Proof. Exercise. By combining Propositions A5.6.7, A5.6.9 and A5.6.10 we obtain the following: A5.6.11 Theorem. space and F a uniform space. A subset G of C(E, F ) is k-compact if (Ascoli’s Theorem) Let (E, τ E) be a topological (a) G is k-closed in C(E, F ), (b) the closure of the set {g(x) : g ∈ G} is compact, for each x ∈ E, and (c) G is equicontinuous. A5.6.12 Remark. If E is a locally compact Hausdorﬀ space and F is a Hausdorﬀ uniform space then the converse of Theorem 5.6.11 is valid; that is, any k-compact subset of G of C(E, F ) satisﬁes conditions (a), (b) and (c). (See Kelley [233]) 555 Exercises A5.6 1. Let (X, U) be any uniform space and (X, τ ) the associated topological space. Show that (X, τ ) is a regular space. 2. If (G, τ ) is a topological group show that the topologies associated with the left uniformity on G, the right uniformity on G, and the two-sided uniformity on G coincide with τ . 3. (i) Let G be a topological group and {Un : n = 1, 2, . . . } a base for the left uniformity on G such that (a) ∞ n=1 Un = diagonal of G × G, (b) Un+1Un+1Un+1 ⊆ Un, and (c) Un = U −1 n , for each n. Show that there exists a metric d on G such that Un ⊆ {(x, y) : d(x, y) < 2−n} ⊆ Un−1, for each n > 1. [Hint: Deﬁne a real-valued function f on G × G by letting f (x, y) = 2−n if (x, y) ∈ Un−1\Un and f (x, y) = 0 if (x, y) belongs to each Un. The desired metric d is constructed from its “ﬁrst approximation”, f , by a chaining argument. For each x and y in G let d(x, y) be the inﬁmum of n i=0 over all ﬁnite sequences x0, x1, . . . , xn+1 such that x = x0 f (xi, xi+1) and y = xn+1.] (ii) Prove that a Hausdorﬀ topological group is metrizable if and only if it satisﬁes the ﬁrst axiom of countability at the identity; that is, there is a countable base of neighbourhoods at the identity. 556 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 4. Let E be a topological space, F a uniform space and G a subset of C(E, F ). Show that (i) if G is equicontinuous at x ∈ E, then the p-closure in F E of G is also equicontinuous at x; and (ii) if G is an equicontinuous subset of C(E, F ), then the p-topology on G is jointly continuous. A5.7 Dual Groups We are now ready to begin our study of duality. If G is an abelian topological group then a continuous A5.7.1 Deﬁnitions. homomorphism γ : G → T is said to be a character. The collection of all characters is called the character group or dual group of G, and is denoted by G∗ or Γ. Observe that G∗ is an abelian group if for each γ1 and γ2 in G∗ we deﬁne (γ1 + γ2)(g) = γ1(g) + γ2(g), for all g ∈ G. Instead of writing γ(g), γ ∈ Γ and g ∈ G we shall generally write (g, γ). Consider the group Z. Each character γ of Z is determined A5.7.2 Example. by γ(1), as γ(n) = nγ(1), for each n ∈ Z. Of course γ(1) can be any element of T. For each a ∈ T, let γa denote the character γ of Z with γ(1) = a. Then the mapping a → γa is clearly an algebraic isomorphism of T onto the character group of Z. So the dual group Z∗ of Z is algebraically isomorphic to T. 557 A5.7.3 Examples. (i) Consider the group T. We claim that every character γ of T can be expressed in the form γ(x) = mx, where m is an integer characterizing the homomorphism γ. To see this let K denote the kernel of γ. Then by Corollary A5.5.6, K = T or K is a ﬁnite cyclic group. If K = T, then γ is the trivial character and γ(x) = 0.x, x ∈ T. If K is a ﬁnite cyclic group of order r then, by Corollart A5.5.5, T/K is topologically isomorphic to T. Indeed, if p is the canonical map of T onto T/K then the topological isomorphism θ : T/K → T is such that θp(x) = rx. Let α be the continuous one-one homomorphism of T into T induced by γ. T γ ....................................................................................................................................................................................................................................................................... ............ .............................................................................................................................................................................................................................................................................................................................................................................................. α ............ ............ T/K T p ................................................................................................................. ................................................................................................................. θ ............ T (ii) Consider the group R. We claim that every character γ of R can be expressed in the form γ(x) = exp(2πidx), x ∈ R, where d is a ﬁxed real number deﬁning γ : γ = γd. Further, we have γa + γb = γa+b. Thus the
dual group of R is algebraically isomorphic to R itself, under the isomorphism d → γd. To prove the claim, let K denote the kernel of γ. If K = R then Proposition A5.5.2 says that K is isomorphic to Z. Further by Corollary A5.5.5 the quotient group R/K is topologically isomorphic to T. As in (i) above there are only two possibilities for the induced algebraic isomorphism R/K → T; these give rise to the cases γ = γ1/a and γ = γ(−1/a), where a is the least element of K. So every γ is of the form γd for some d ∈ R, and hence R is algebraically isomorphic to its dual group. If K = R then γ = γ0. Exercises A5.7 #1 implies that α(x) = x, for all x ∈ T, or α(x) = −x, for all x ∈ T. So γ(x) = rx or −rx, for each x ∈ T. 558 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Hence each character γ of T is of the form γ = γm for some m ∈ Z, where γm(x) = mx for all x ∈ T. Of course γm + γn = γm+n. Thus the dual group T∗ of T is algebraically isomorphic to Z, with the isomorphism being m → γm. We now topologize G∗. A5.7.4 Remark. Note that G∗ is a p-closed subset of C(G, T). Let G be any abelian topological group. Then G∗ A5.7.5 Proposition. endowed with the p-topology or the k-topology is a Hausdorﬀ abelian topological group. Proof. Exercise. A5.7.6 Theorem. k-topology, is an LCA-group. If G is any LCA-group then G∗, endowed with the 559 To show that G∗, with the k-topology, is locally compact, let U be any Proof. compact neighbourhood of 0 in G and Va = {t : t = exp(2πix) ∈ T and 1 > x > 1 − a or a > x 0}, where a is a positive real number less than 1 4 . Then Va is an open neighbourhood of 0 in T. Let Na = P (U, Va) = {γ ∈ G∗ : (g, γ) ∈ Va, for each g ∈ U }. By the deﬁnition of the k-topology, Na is a neighbourhood of 0 in G∗. we shall show that the k-closure of Na, clk(Na), is k-compact. To do this we use Ascoli’s Theorem A5.6.11. Firstly, we show that Na is equicontinuous. Let ε > 0 be given. We wish to show that there exists a neighbourhood U1 of 0 in G such that for all γ ∈ Na and g, h, and g − h in U1, (g − h, γ) = (g, γ) − (h, γ) ∈ Vε, where Vε = {t : t = exp(2πix) ∈ T and 1 > x > 1 − ε or ε > x 0}. n Suppose that there is no such U1. Without loss of generality assume ε < 1 4 and 2 > nε > a. Further, let W be a neighbourhood let n be a positive integer such that 1 of 0 in G such that Wi ⊆ U, where each Wi = W. (1) i=1 By assumption, then, for some g and h in W with g − h ∈ W and some γ ∈ Na, (g − h, γ) ∈ Vε. So without loss of generality (g − h, γ) = exp(2πix) with a > x ε. Let j be a positive integer less than or equal to n such that 1 2 > jx > a. So (j(g − h), γ) ∈ Va. But as jg, jh and j(g − h) all belong to U , by (1), (j(g − h), γ) ∈ Va, which is a contradiction. Hence Na is equicontinuous. By Proposiition A5.6.9 the p-closure of Na is equicontinuous. As any subset of an equicontinuous set is equicontinuous, and clk(Na) is a subset of the p-closure of Na, we have that clk(Na) is equicontinuous. As T is compact, condition (b) of Ascoli’s Theorem A5.6.11 is also satisﬁed and hence clk(Na) is a compact neighbourhood of 0. So G∗ with the k-topology is locally compact. 560 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE As a corollary to the proof of Theorem A5.7.6 we have Let G be any LCA-group, Γ its dual group endowed with A5.7.7 Corollary. the k-topology, K a compact neighbourhood of 0 in G and for some positive real number a < 1 4, Va = {t : t = exp(2πix) ∈ T with 1 > x > 1 − a or a > x 0}. Then P (K, Va) is a compact neighbourhood of 0 in Γ. A5.7.8 Notation. with the k-topology. From now on G∗ and Γ will denote the dual group of G A5.7.9 Theorem. compact, then Γ is discrete. If G is discrete, then Γ is compact. Let G be an LCA-group and Γ its dual group. If G is Let G be compact and Va be as in Corollary A5.7.7 Then P (G, Va) is a Proof. neighbourhood of 0 in Γ. As Va contains no subgroup other than {0}, we must have P (G, Va) = {0}. So Γ has the discrete topology. Let G be discrete. Then by Corollary A5.7.7, P ({0}, Va) is a compact subset of Γ. But P ({0}, Va) clearly equals Γ, and hence Γ is compact. A5.7.10 Corollary. Z. The dual group T∗ of T is topologically isomorphic to Exercises A5.7 1. Show that if γ is a continuous one-to-one homomorphism of T into itself then either γ(x) = x, for all x ∈ T or γ(x) = −x, for all x ∈ T. [Hint: Firstly show that γ must be onto. Next, observe that T has only one element of order 2.] 2. Show that if G is any abelian topological group, then G∗ endowed with the p-topology or the k-topology is a Hausdorﬀ topological group. [Hint: Let γ1 − γ2 ∈ P (K, U ). Let W be an open symmetric neighbourhood of 0 in T such that 2W + (γ1 − γ2)(K) ⊆ U . Observe that 561 [γ1 + P (K, W )] − [γ2 + P (K, W )] ⊆ P (K, U ).] 3. Show that the dual group of Z is topologically isomorphic to T. 4. Show that R is topologically isomorphic to its dual group. 5. Find the dual groups of the discrete ﬁnite cyclic groups. 6. Let G be any abelian topological group and G∗ its dual group. Show that the family of all sets P (K, Vε), as K ranges over all compact subsets of G containing O and ε ranges over all positive numbers less than one, is a base of open neighbourhoods of O for the k-topology on G∗. 562 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.8 Pontryagin–van Kampen Duality Theorem: Introduction We begin with a statement of the duality theorem. Theorem. [Pontryagin-van Kampen Duality Theorem] Let G be an LCAgroup and Γ its dual group. For ﬁxed g ∈ G, let g : Γ → T be the function given by g(γ) = γ(g), for all γ ∈ Γ. If α : G → Γ∗ is the mapping given by α(g) = g, then α is a topological group isomorphism of G onto Γ∗. A5.8.1 Remarks. (i) Roughly speaking this says that every LCA-group is the dual group of its dual group. (ii) This theorem says that every piece of information about an LCA-group is contained in some piece of information about its dual group. In particular all information about a compact Hausdorﬀ abelian group is contained in information about its dual group – a discrete abelian group. So any compact Hausdorﬀ abelian group can be completely described by the purely algebraic properties of its dual group; for example, if G is a compact Hausdorﬀ abelian group then we shall see that (a) G is metrizable if and only if Γ is countable. (b) G is connected if and only if Γ is torsion-free. A5.8.2 Lemma. homomorphism of G into Γ∗. In the notation of the above Theorem, α is a continuous 563 Firstly we have to show that α(g) ∈ Γ∗; that is, α(g) = g is a continuous Proof. homomorphism of Γ into T. As α(g)(γ1 + γ2) = (γ1 + γ2)(g) = γ1(g) + γ2(g) = α(g)(γ1) + α(g)(γ2), for each γ1 and γ2 in Γ, α(g) : Γ → T is a homomorphism. To see that α(g) is continuous, it suﬃces to note that α(g)(γ) ∈ Vε whenever γ ∈ P ({g}, Vε), where Vε is an ε-neighbourhood of 0 in T as in Theorem A5.7.6. So α is a map of G into Γ∗. That α is a homomorphism follows by observing α(g1 + g2)(γ) = γ(g1 + g2) = γ(g1) + γ(g2) = α(g1)(γ) + α(g2)(γ), for all γ ∈ Γ, =⇒ α(g1 + g2) = α(g1) + α(g2), for all g1, g2 ∈ G. To show that α is continuous, it suﬃces to verify continuity at 0 ∈ G. Let W be any neighbourhood of 0 in Γ∗. Without loss of generality we can assume W = P (K, Vε), for some compact subset K of Γ. We have to ﬁnd a neighbourhood of 0 in G which maps into W . Let U be any open neighbourhood of 0 in G such that U is compact and consider of 0 in Γ. The collection the neighbourhood P covers the compact set K and so there exist γ1, . . . , γm in Γ such that U , Vε/2 U , Vε/2 : γ ∈ Γ γ + P K ⊆ γ1 + P U , Vε/2 ∪ · · · ∪ γm + P U , Vε/2 . Let U1 be a neighbourhood of 0 in G such that U1 ⊆ U and γi(g) ∈ Vε/2, for all g ∈ U1 and i = 1, . . . , m. (This is possible since the γi are continuous.) We claim that U1 is the required neighbourhood. To see this let g ∈ U1 and consider , for some i ∈ {1, . . . , m}. So α(g)(γ), where γ ∈ K. Then γ ∈ γi + P γ − γi ∈ P this implies that γ(g) ∈ Vε/2 + Vε/2 ⊆ Vε. So α(g)(γ) ∈ Vε, as required. . Thus (γ − γi)(g) ∈ Vε/2 for g ∈ U1 ⊆ U . As γi(g) ∈ Vε/2, U , Vε/2 U , Vε/2 564 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE The duality theorem will be proved for compact groups and discrete groups ﬁrst, and then it will extended to all LCA-groups. There are a number of proofs of the duality theorem in the literature. The proof presented in this Appendix is from Morris [293]. An elegant proof appears in Rudin [353]. Hewitt and Ross [180] present the more classical approach of ﬁrst deriving the structure theory of LCA-groups and then using it in the proof of duality. A proof using category theory is given in Roeder [343]. Other references include Weil [421], Cartan and Godement [71], Raikov [335], Naimark [306], Dikranjan et al. [108] and of course, Pontryagin [330]. Exercises A5.8 1. Show that Z satisﬁes the Pontryagin-van Kampen Duality Theorem. (Note. This requires more than just showing that Z∗∗ is topologically isomorphic to Z. You must prove that the map α in the duality theorem is a topological group isomorphism.) [Hint. See Example A5.7.2 and Example A5.7.3.] 2. Show that T satisﬁes the Pontryagin-van Kampen Duality Theorem. [Hint. Firstly show that α is 1-1 and onto. Then use the Open Mapping Theorem A5.4.4. 3. Prove that every discrete ﬁnite cyclic group satisﬁes the Pontryagin-van Kampen Duality Theorem. 4. Prove that the topological group R satisﬁes the Pontryagin-van Kampen Duality Theorem. A5.9 Dual Groups of Subgroups, Quotients, and Finite Products 565 Wenow make some observations which are needed in the proof of the duality theorem, but which are also of interest in themselves. If G1, . . . , Gn are LCA-groups with dual groups A5.9.1 Theorem. Γ1, . . . , Γn, respectively, then Γ = Γ1 × Γ2 × · · · × Γn is the dual group of G1 × G2 × · · · × Gn. Proof. It suﬃces to prove this for the case n = 2, and ﬁnite products can then be easily deduced by mathematical induction. If g = g1 + g2 is the unique representation of g ∈ G as a sum of elements of G1 and G2, then the pair γ1 ∈ Γ1
and γ2 ∈ Γ2 determine a character γ ∈ Γ by the formula (g, γ) = (g1, γ1) + (g2, γ2) (1) Since every γ ∈ Γ is completely determine by its action on the subgroups G1 and G2, equation (1) shows that Γ is algebraically the direct sum of Γ1 and Γ2. To see that Γ has the product topology Γ1 × Γ2 simply note that (a) P (K, Vε) ⊇ P K1, Vε/2 , where K is any compact subset of G = G1 × G2, K1 = p1(K), K2 = p2(K) and p1 and p2 are the projections of G onto G1 and G2, respectively, and K2, Vε/2 + P (b) if K1 is a compact subset of G1 containing 0 and K2 is a compact subset of G2 containing 0, then P (K1 × K2, Vε) ⊆ P (K1, Vε) + P (K2, Vε) . (See Exercises A5.7 #6.) A5.9.2 Corollary. For each n 1, Rn is topologically isomorphic to its dual group. 566 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.9.3 Corollary. For each n 1, the topological groups Tn and Zn are dual groups of each other. If G1, . . . , Gn are LCA-groups which satisfy the A5.9.4 Corollary. Pontryagin-van Kampen Duality Theorem, then G1 × G2 × · · · × Gn satisﬁes the Pontryagin-van Kampen Duality Theorem. Hence Ra × Tb × G satisﬁes the Pontryagin-van Kampen Duality Theorem, where G is a discrete ﬁnitely generated abelian group, and a and b are non-negative integers. Proof. Exercise. Theorem A5.9.1 shows that the dual group of a ﬁnite product is the product of the dual groups. We shall see, in due course, that the dual of a closed subgroup is a quotient group, and the dual of a quotient group is a closed subgroup. As a ﬁrst step towards this we have Proposition A5.9.5. Let f be a continuous homomorphism of an LCAA5.9.5 Proposition. group A into an LCA-group B. Let a map f ∗ : B∗ → A∗ be deﬁned by putting f ∗(γ)(a) = γf (a), for each γ ∈ B∗ and a ∈ A. Then f ∗ is a continuous homomorphism of B∗ into A∗. If f is onto, then f ∗ is one-one. If f is both an open mapping and one-one, then f ∗ is onto. 567 A f ................................................................................................................................................................... ............ ...................................................................................................................................................... ............ f ∗(γ) γ B ................................................................................................................................................................... ............ T The veriﬁcation that f ∗ is a homomorphism of B∗ into A∗ is routine. To Proof. see that f ∗ is continuous, let P (K, U ) be a subbasic open set in A∗, where U is an open subset of T and K is a compact subset of A. The continuity of f ∗ follows from the fact that (f ∗)−1(P (K, U )) = P (f (K), U ) is an open subset of B∗. Assume f is onto and suppose that f ∗(γ1) = f ∗(γ2) where γ1 and γ2 are in B∗. Then f ∗(γ1)(a) = f ∗(γ2)(a), for all a ∈ A; that is, γ1f (a) = γ2f (a), for all a ∈ A. As f is onto this says that γ1(b) = γ2(b), for all b ∈ B. Hence γ1 = γ2 and f ∗ is one-one. Assume that f is both an open mapping and one-one. Let δ ∈ A∗. As f is one-one, Proposition A5.3.6 tells us that there is a (not necessarily continuous) homomorphism γ : B → T such that δ = γf . As δ is continuous and f is an open mapping, γ is indeed continuous; that is, γ ∈ B∗. As f ∗(γ) = δ, we have that f ∗ is onto. A5.9.6 Corollary. If B is a quotient group of A, where A and B are either both compact Hausdorﬀ abelian groups or both discrete abelian groups, then B∗ is topologically isomorphic to a subgroup of A∗. Proof. Exercise. 568 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.9.7 Corollary. abelian groups, then A∗ is a quotient group of B∗. If A is a subgroup of B, where A and B are discrete Proof. Exercise. A5.9.8 Remark. As noted earlier we shall see in due course that Corollary A5.9.6 and Corollary A5.9.7 remain true if the hypotheses “compact Hausdorﬀ" and “discrete" are replaced by “locally compact Hausdorﬀ". We also mention that the analogous results for duality in Banach spaces appears in Exercises 10.3 #33(xvi). The next lemma indicates that, before proving the Pontryagin-van Kampen Duality Theorem, we shall have to see that LCA-groups have enough characters to separate points. A5.9.9 Lemma. In the notation of the Pontryagin-van Kampen Duality Theorem, the map α is one-one if and only if G has enough characters to separate points; that is, for each g and h in G, with g = h, there is a γ ∈ Γ such that γ(g) = γ(h). Assume that α is one-one. Suppose that there exist g and h in G, with Proof. g = h, such that γ(g) = γ(h) for all γ ∈ Γ. Then α(g)(γ) = α(h)(γ), for all γ ∈ Γ. So α(g) = α(h), which implies that g = h, a contradiction. Hence G has enough characters to separate points. Assume now that G has enough characters to separate points. Let g and h be in G, with g = h. Then there is a γ ∈ Γ such that γ(g) = γ(h). So α(g)(γ) = α(h)(γ), which implies that α(g) = α(h). So α is one-one. The ﬁnal proposition in this section should remind some readers of the Stone- Weierstrass Theorem. 569 Let G be an LCA-group and Γ its dual group. A5.9.10 Proposition. Let G satisfy the Pontryagin-van Kampen Duality Theorem and also have the property that every nontrivial Hausdorﬀ quotient group Γ/B of Γ has a nontrivial character. If A is a subgroup of Γ which separates points of G, then A is dense in Γ. Proof. Suppose A is not dense in Γ. If B is the closure of A in Γ then Γ/B is a nontrivial LCA-group. So there exists a nontrivial continuous homomorphism φ : Γ/B → T. Let f be the canonical homomorphism : Γ → Γ/B. Then φf is a continuous homomorphism : Γ → T. Furthermore, φf (Γ) = 0 but φf (B) = 0. As G satisﬁes the Pontryagin-van Kampen Duality Theorem, there is a g ∈ G such that φf (γ) = γ(g), for all γ ∈ Γ. So γ(g) = 0 for all γ in A. But since A separates points in G, this implies g = 0. So φf (Γ) = 0, which is a contradiction. Hence A is dense in Γ. Of course the second sentence in the statement of Proposition A5.9.10 will, in due course, be seen to be redundant. 570 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Exercises A5.9 1. (i) Show that if G1, . . . , Gn are LCA-groups which satisfy the Pontryagin-van Kampen Duality Theorem, then G1 × G2 × · · · × Gn satisﬁes the Pontryaginvan Kampen Duality Theorem. (ii) Deduce that ever discrete ﬁnitely generated abelian group satisﬁes the Pontryagin-van Kampen Duality Theorem. [Hint: Use the fact that every ﬁnitely generated abelian group is a direct product of a ﬁnite number of cyclic groups.] (iii) Hence show that Ra × Tb × G satisﬁes the Pontryagin-van Kampen Duality Theorem, where G is a discrete ﬁnitely generated abelian group, and a and b are non-negative integers. 2. Show that if B is a quotient group of A, where A and B are either both compact Hausdorﬀ abelian groups or both discrete abelian groups, then B∗ is topologically isomorphic to a subgroup of A∗. 3. Show that if A is a subgroup of B, where A and B are discrete abelian groups, then A∗ is a quotient group of B∗. 4. Show that if G is any LCA-group and Γ is its dual group, then Γ has enough characters to separate points. A5.10 Peter-Weyl Theorem 571 In Lemma A5.9.9 we saw that a necessary condition for a topological group to satisfy duality is that it have enough characters to separate points. That discrete abelian groups have this property has been indicated already in Corollary A5.3.7. For compact groups we use a result from the representation theory of topological groups. [For a brief outline of this theory, see Higgins [182]. Fuller discussions appear in Adams [5], Hewitt and Ross [180], Pontryagin [330] and Hofmann [192].] The following theorem is named after Hermann Weyl (1885–1955) and his student Fritz Peter (1899âĂŞ-1949). Weyl and Peter proved this in Weyl and Peter [422] in 1927 for compact Lie groups and by the Dutch mathematician Egbert Rudolf van Kampen (1908–1942)extended it to all compact groups in van Kampen [397] in 1935 using the 1934 work, von Neumann [411], of the Hungarian-American mathematician John von Neumann (1903–1957) on almost periodic functions. Hausdorﬀ group. A5.10.1 Theorem. (Peter-Weyl Theorem) Let G be a compact Then G has suﬃciently many irreducible continuous In other words, for each g ∈ G, g = e, representations by unitary matrices. there is a continuous homomorphism φ of G into the unitary group U (n), for some n, such that φ(g) = e. If G is abelian then, without loss of generality, it can be assumed that n = 1. As U (1) = T we obtain the following corollary, which is the result which we shall use in proving duality for compact abelian groups. A5.10.2 Corollary. Every compact Hausdorﬀ abelian topological group has enough characters to separate points. Corollary A5.10.2 was ﬁrst proved by John von Neumann for compact metrizable abelian groups. A derivation of Corollary A5.10.2 from von Neumann’s result is outlined in Exercises A5.10 #2 and #3. 572 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.10.3 Corollary. G is topologically isomorphic to a closed subgroup of the product i∈I each Ti is topologically isomorphic to T, and I is some index set. Let G be any compact Hausdorﬀ abelian group. Then Ti, where Proof. Exercise. Let G be a compact Hausdorﬀ abelian group. Then A5.10.4 Corollary. every neighbourhood U of 0 contains a closed subgroup H such that G/H is topologically isomorphic to Tn × D, for some ﬁnite discrete group D and n 0. Proof. Exercise. Exercises A5.10 1. Using Exercises A5.3 #4(i), show that every compact totally disconnected abelian topological group has enough characters to separate points. 2. Show that every compactly generated locally compact Hausdorﬀ group G can be approximated by metrizable groups in the following sense: For each neighbourhood U of e, there exists a compact normal subgroup H of G such that H ⊆ U and G/H is metrizable. [Hint: Let V1, V2, V3, . . . be a sequence of symmetric compact neighbourhoods of e such that (i) V1 ⊆ U , (ii) V 2 n+1 ⊆ Vn, for n 1, and (iii) g−1Vng
⊆ Vn−1, for n 2 and g ∈ K, where K is a compact set which generates G. Put H = Vn and use Exercises A5.6 #3.] ∞ n=1 3. Using Exercise 2 above, deduce statement B from statement A. (A) Every compact metrizable abelian group has enough characters to separate points. 573 (B) Every compact Hausdorﬀ abelian group has enough characters to separate points. 4. (i) Show that every compact Hausdorﬀ abelian group G is topologically isomorphic to a subgroup of a product i∈I Ti of copies of T. [Hint: See the proof of Theorem A5.3.9.] (ii) If G is also metrizable show that the index set I can be chosen to be a countable set. [Hint: Use Exercises A5.6 #3(ii).] (iii) Using (i) show that if G is any compact Hausdorﬀ abelian group, then every neighbourhood U of 0 contains a closed subgroup H such that G/H is topologically isomorphic to Tn × D, for some ﬁnite discrete group D and n 0. [Hint: Reread Remark A5.3.1. Use Corollary A5.5.20.] 5. Show that every compact Hausdorﬀ group is topologically isomorphic to a subgroup of a product of copies of U , where U = U (n). ∞ n=1 574 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.11 The Duality Theorem for Compact Groups and Discrete Groups The next proposition provides the last piece of information we need in order to prove the Pontryagin-van Kampen Duality Theorem for compact groups and discrete groups. (This proposition should be compared with Proposition A5.9.10.) Let G be a discrete abelian group and Γ its dual A5.11.1 Proposition. group. If A is a subgroup of Γ which separates points of G, then A is dense in Γ. Noting how the topology on Γ is deﬁned, it suﬃces to show that each Proof. non-empty sub-basic open set P (K, U ), where K is a compact subset of G and U is an open subset of T, intersects A nontrivially. As G is discrete, K is ﬁnite. Let H be the subgroup of G generated by K and f ∗ : Γ → H∗ the map obtained by restricting the characters of G to H. According to Proposition A5.9.5 and Corollary A5.9.7, f ∗ is an open continuous homomorphism of Γ onto H∗. As A separates points of G, f ∗(A) separates points of H. Observing that Corollary 5.9.4 says that H satisﬁes the Pontryagin-van Kampen Duality Theorem, Proposition A5.9.10 then implies that f ∗(A) is dense in H∗. So f ∗(P (K, U )) ∩ f ∗(A) = Ø. In other words there is a γ ∈ A such that, when restricted to H, γ maps K into U . Of course this says that γ ∈ P (K, U ) ∩ A. A5.11.2 Theorem. (Pontryagin-van Kampen Duality Theorem for Compact Groups) Let G be a compact Hausdorﬀ abelian group and Γ its dual group. Then the canonical map α of G into Γ∗ is a topological group isomorphism of G onto Γ∗. 575 By Lemma A5.8.2, Lemma A5.9.9 and Theorem A5.10.1, α is a Proof. continuous one-one homomorphism of G into Γ∗. Clearly α(G) separates points of Γ. As Γ is discrete, Proposition A5.11.1 then implies that α(G) is dense in Γ∗. However, α(G) is compact and hence closed in Γ∗. Thus α(G) = Γ∗; that is, the map α is onto. Finally, the Open Mapping Theorem for Locally Compact Groups A5.4.4 tells us that α is also an open map. A5.11.3 Corollary. dual group. If A is a subgroup of Γ which separates points of G, then A = Γ. Let G be a compact Hausdorﬀ abelian group and Γ its Proof. This is an immediate consequence of Theorem A5.11.2, Proposition A 5.9.10, Corollary A5.3.7, and Corollary A5.7.7. Let G be an LCA-group with enough characters to A5.11.4 Corollary. separate points and K a compact subgroup of G. Then every character of K extends to a character of G. The collection of characters of K which extend to characters of G form Proof. a subgroup A of K∗. As G has enough characters to separate points, A separates points of K. So by Corollary A5.11.3, A = K∗. Let B be an LCA-group with enough characters to A5.11.5 Corollary. separate points and f a continuous one-one homomorphism of a compact group A into B. Then the map f ∗ : B∗ → A∗, described in Proposition A5.9.5, is a quotient homomorphism. 576 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Proof. Exercise. A5.11.6 Theorem. (Pontryagin-van Kampen Duality Theorem for Discrete Groups) Let G be a discrete abelian group and Γ its dual group. Then the canonical map α is a topological group isomorphism of G onto Γ∗. As in Theorem A5.11.2, α is a continuous one-one homomorphism of G Proof. into Γ∗. As α(G) separates points of Γ and Γ is compact, Corollary A5.11.3 yields that α(G) = Γ. As G and Γ∗ are discrete this completes the proof. We conclude this section by showing how duality theory yields a complete description of compact Hausdorﬀ abelian torsion groups. (Recall that a group G is said to be a torsion group if each of its elements is of ﬁnite order.) The ﬁrst step is the following interesting result. A5.11.7 Deﬁnition. set I. Then the subgroup H = i∈I Let Gi, i ∈ I, be a set of groups, for some index i∈I Gi, where r Gi of the direct product g = (. . . , gi, . . . ) ∈ H if and only if each gi ∈ Gi and gi is the identity element for all but a ﬁnite number of i ∈ I, is said to be the restricted direct product. If G = i∈I Gi is the direct product of any family A5.11.8 Theorem. {Gi : i ∈ I} of compact Hausdorﬀ abelian groups, then the discrete group G∗ is algebraically isomorphic to the restricted direct product r Γi of the i∈I corresponding dual groups {Γi = G∗ i : i ∈ I}. 577 Each g ∈ G may be thought of as a “string" g = (. . . , gi, . . . ), the Proof. group operation being componentwise addition. If γ = (. . . , γi, . . . ), where γi ∈ Γi and only ﬁnitely many γi are non-zero, then γ is a character on G deﬁned by (g, γ) = i∈I (gi, γi), for each g ∈ G. (Observe that this is a ﬁnite sum!) Let us denote the subgroup of G∗ consisting of all such γ by A. Then A is algebraically isomorphic to the restricted direct product i∈I We claim that A separates points of G. To see this let g ∈ G, g = 0. Then g = (. . . , gi, . . . ) with some gi = 0. So there is a γi ∈ Γi such that γi(gi) = 0. Putting γ = (. . . , γj, . . . ) where γj = 0 unless j = i, we see that γ(g) = γi(gi) = 0. As γ ∈ A, A separates points of G. By Corollary A5.11.3, this implies that A = G∗. r Γi. A5.11.9 Corollary. Every countable abelian group is algebraically isomorphic to a quotient group of a countable restricted direct product of copies of Z. Let G be a countable abelian group. Put the discrete topology on G and Proof. let Γ be its dual group. Of course Γ is compact and by Exercises A5.10 #4, Γ is topologically isomorphic to a subgroup of a product Ti of copies of T, where the i∈I cardinality of the index set I equals the cardinality of Γ∗. By Theorem A5.11.6, Γ∗ is topologically isomorphic to G. So Γ is topologically isomorphic to a subgroup of a countable product of copies of T. Taking dual groups and using Theorem A5.11.8 and Corollary A5.11.5 we obtain the required result. 578 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Of course the above corollary also follows from the fact that the free abelian group on a countable set is a countable restricted direct product of copies of Z, and that any countable abelian group is a quotient group of the free abelian group on a countable set. A5.11.10 Remark. Kaplan [227, 228] has investigated generalizations of Theorem A5.11.7 to direct products of non-compact groups. As a direct product of LCA-groups is not, in general, an LCA-group we must ﬁrst say what we mean by the dual group of a non-LCA-group: If G is any abelian topological group we deﬁne Γ to be the group of continuous homomorphisms of G into T, with the compact open topology. Then Γ is an abelian topological group and we can form Γ∗ in the same way. As in the locally compact case there is a natural map α which takes g ∈ G to α(g) a function from Γ into T. We can then ask for which groups is α a topological group isomorphism of G onto Γ∗. Such groups will be called reﬂexive. A satisfactory description of this class is not known, but it includes not only all LCA-groups but also all Banach spaces (considered as topological groups). (See Smith [366].) Kaplan showed that if {Gi : i ∈ I} is a family of reﬂexive groups then Gi is also a reﬂexive group. Its dual group is algebraically isomorphic to the i∈I restricted direct product of the family {G∗ i : i ∈ I}. The topology of the dual group is slightly complicated to describe, but when I is countable it is simply the subspace topology induced on i is given the box topology. In particular this is i∈I i if i∈I rG∗ G∗ the case when each Gi is an LCA-group–thus generalizing Theorem A5.11.7. For further comments on reﬂexive groups see Brown et al. [60], Venkataraman [403], Noble [318], Varopoulos [401], and Vilenkin [405]. Reﬂexivity of abelian topological groups, including nuclear groups, has been thoroughly studied in Banaszczyk [30]. For relevant research on reﬂexivity, see Hofmann and Morris [190], Nickolas [316], Nickolas [315],Chasco et al. [79], Chasco and Martin-Peinador [78], Chasco and DomÃŋnguez [77], Chasco [76], Barr [34] and Pestov [324]. To prove the structure theorem of compact Hausdorﬀ abelian torsion groups we have to borrow the following result of abelian group theory. (See Fuchs [143].) A5.11.11 Theorem. bounded order is algebraically isomorphic to a restricted direct product i∈I An abelian group all of whose elements are of rZ(bi), with only a ﬁnite number of the bi distinct, where Z(bi) is the discrete cyclic group with bi elements. 579 A5.11.12 Theorem. topologically isomorphic to i∈I only a ﬁnite number of distinct bi. A compact Hausdorﬀ abelian torsion group is Z(bi), where I is some index set and there exist Proof. Exercise. Exercises A5.11 1. If f is a continuous one-one homomorphism of a compact group A into an LCAgroup B which has enough characters to separate points, show that the map f ∗ : B∗ → A∗, described in Proposition A5.9.5, is a quotient homomorphism. 2. Show that every compact Hausdorﬀ abelian torsion group G is topologically Z(bi), where Z(bi) is a discrete cyclic group with bi isomorphic to a product i∈I elements, I is an index set,
and there are only a ﬁnite number of distinct bi. G(n) and, using the [Hint: Let G(n) = {x ∈ G : nx = 0}. Observe that G = Baire Category Theorem A5.4.1, show that one of the quotient groups G/G(n) is ﬁnite. ∞ n=1 Deduce that the orders of all elements of G are bounded. Then use the structure theorem of abelian groups of bounded order A5.11.11.] 580 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.12 Monothetic LCA-groups and Compactly Generated LCA-groups We have proved the duality theorem for compact groups and for discrete groups. To extend the duality theorem to all LCA-groups we shall prove two special cases of the following proposition: If G is an LCA-group with a subgroup H such that both H and G/H satisfy the duality theorem, then G satisﬁes the duality theorem. The two cases we prove are when H is compact and when H is open. The duality theorem for all LCA-groups then follows from the fact that every LCAgroup G has an open subgroup H which in turn has a compact subgroup K such that H/K is an “elementary group" which is known to satisfy the duality theorem. By an “elementary group" we mean one which is of the form Ra × Zb × Tc × F , where F is a ﬁnite discrete abelian group and a, b and c are non-negative integers. Once we have the duality theorem we use it, together with the above structural result, to prove the Principal Structure Theorem. We begin with some structure theory. A5.12.1 Deﬁnition. A topological group is said to be monothetic if it has a dense cyclic subgroup. A5.12.2 Examples. Z and T are monothetic. A5.12.3 Theorem. compact or G is topologically isomorphic to Z. Let G be a monothetic LCA-group. Then either G is 581 If G is discrete then either G = Z or G is a ﬁnite cyclic group and hence Proof. is compact. So we have to prove that G is compact if it is not discrete. Assume G is not discrete. Then the dense cyclic subgroup {xn : n = 0, ±1, ±2, . . . }, where xn + xm = xn+m for each n and m, is inﬁnite. (If the cyclic subgroup were ﬁnite it would be discrete and hence closed in G. As it is also dense in G, this would mean that it would equal G and G would be discrete.) Let V be an open symmetric neighbourhood of 0 in G with V compact. If g ∈ G, then V + g contains some xk. So there is a symmetric neighbourhood W of 0 in G such that (g − xk) + W ⊆ V . As G is not discrete, W contains an inﬁnite number of the xn and as W is symmetric x−n ∈ W if xn ∈ W . Hence there exists a j < k such that xj ∈ W . Putting i = k − j we have i > 0 and g − xi = g − xk + xj ∈ g − xk + W ⊆ V. This proves that G = ∞ i=1 (xi + V ). (The important point is that we need xi only for i > 0 .) As V is a compact subset of G we have that N V ⊆ (xi + V ), for some N. i=1 (1) For each g ∈ G, let n = n(g) be the smallest positive n such that g ∈ xn + V . By (1), xn − g ∈ xi + V for some 1 i N . So we have that g ∈ xn−i + V . Since i > 0, n − i < n and so by our choice of n, n − i 0. Thus n i N . So for each g ∈ G, n N , which means that N G = (xi + V ) i=1 which is a ﬁnite union of compact sets and so G is compact. 582 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A compact Hausdorﬀ abelian group G is monothetic if A5.12.4 Theorem. and only if G∗ is topologically isomorphic to a subgroup of Td, the circle group endowed with the discrete topology. Proof. Exercise. We now use Theorem to obtain our ﬁrst description of the structure of compactly (Recall that an LCA-group G is said to be compactly generated LCA-groups. generated if it has a compact subset V such that G is generated algebraically by V . Without loss of generality V can be chosen to be a symmetric neighbourhood of 0.) If G is an LCA-group which is algebraically A5.12.5 Proposition. generated by a compact symmetric neighbourhood V of 0, then G has a closed subgroup A topologically isomorphic to Zn, for some integer n 0, such that G/A is compact and V ∩ A = {0}. ∞ n=1 m i=1 Proof. n 1, then G = If we put V1 = V and inductively deﬁne Vn+1 = Vn + V , for each integer Vn. As V2 is compact there are elements g1, . . . , gm in G such that V2 ⊆ (gi + V ). Let H be the group generated by {g1, . . . , gm}. So Vi ⊆ V + H, for i = 1 and i = 2. If we assume that Vn ⊆ V + H, then we have Vn+1 ⊆ V + (V + H) = V2 + H ⊆ (V + H) + H = V + H. So, by induction, Vn ⊆ V + H, for all n 1, and hence G = V + H. Let Hi be the closure in G of the subgroup Hi generated by gi, for i = 1, . . . , m. If each Hi is compact, then as H = H1 + · · · + Hm, H is compact and so G = V + H is compact. (Use Exercises A5.1 #4.) The Proposition would then If G is not compact, then, by Theorem A5.12.4, one of the be true with n = 0. monothetic groups Hi is topologically isomorphic to Z. In this case Hi = Hi and we deduce that 583 if G = V +H, where H is a ﬁnitely generated group, and G is not compact, then H has a subgroup topologically isomorphic to Z. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . () As H is a ﬁnitely generated abelian group (and every subgroup of an abelian group with p generators can be generated by p elements) there is a largest n such that H contains a subgroup A topologically isomorphic to Zn. Since A is discrete and V is compact, A ∩ V is ﬁnite. Without loss of generality we can assume (If necessary we replace A by a subgroup A which is also that A ∩ V = {0}. topologically isomorphic to Zn and has the property that A ∩ V {0}. For example, if A = gp {a1, . . . , an} and r is chosen such that A ∩ V ⊆ {k1a1 + · · · + knan : 1 − r ki r − 1, i = 1, . . . , n} then we put A = gp {ra1, . . . , ran}.) Let f be the canonical homomorphism of G onto K = G/A. Then K = f (V ) + f (H). By Exercises A5.12 #2 and our choice of n, f (H) has no subgroup topologically isomorphic to Z. By () applied to K instead of G, we see that K is compact, as required. The above proposition allows us to prove a most important theorem which generalizes Theorem A5.10.1. 584 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.12.6 Theorem. Every LCA-group has enough characters to separate points. Let G be any LCA-group and g any non-zero element of G. Let V be Proof. a compact symmetric neighbourhood of 0 which contains g. Then the subgroup H generated algebraically by V is, by Proposition A5.2.9, an open subgroup of G. By Proposition A5.12.5, H has a closed subgroup A such that H/A is compact and V ∩ H = {0}. Deﬁning f to be the canonical map of H onto H/A we see that f (g) = 0. According to Theorem A5.10.1 there is a continuous homomorphism φ : H/A → T such that φ(f (g)) = 0. Then φf is a continuous homomorphism of H into T. As H is an open subgroup of G and T is divisible, Proposition A5.3.6 tells us that φf can be extended to a continuous homomorphism γ : G → T. Clearly γ(g) = 0 and so G has enough characters to separate points. A5.12.7 Corollary. If g is any element of G not in H, then there is a character γ of G such that γ(g) = 0 but γ(h) = 0, for all h ∈ H. Let H be a closed subgroup of an LCA-group G. Proof. Exercise. The next corollary is an immediate consequence of the opening sentences in the proof of Theorem A5.12.6. A5.12.8 Corollary. Every LCA-group has a subgroup which is both open and a compactly generated LCA-group. 585 A5.12.9 Remarks. Theorem A5.12.6 was ﬁrst proved by E.R. van Kampen. A proof based on the theory of Banach algebras was given by Israil Moiseevic Gelfand and Dmitrii Abramovich Raikov in Gelfand and Raikov [152]. The reader should not be misled, by Theorem A5.12.6, into thinking that all Hausdorﬀ abelian topological groups have enough characters to separate points. This is not so. See §23.32 of Hewitt and Ross [180]. Gelfand 586 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE The next proposition gives another useful description of the structure of compactly generated LCA-groups. If G is a compactly generated LCA-group, then A5.12.10 Proposition. it has a compact subgroup K such that G/K is topologically isomorphic to Ra × Zb × Tc × F , where F is a ﬁnite discrete abelian group and a, b and c are non-negative integers. Proof. By Proposition A5.12.5 there exists a discrete ﬁnitely generated subgroup D of G such that G/D is compact. Let N be a compact symmetric neighbourhood of 0 such that 3N ∩ D = {0}. If f : G → G/D is the canonical homomorphism then f (N ) is a neighbourhood of 0 in G/D and, by CorollaryA5.12.8, there exists a closed subgroup B ⊆ f (N ) such that (G/D)/B is topologically isomorphic to Tn × E, where E is a ﬁnite discrete group and n 0. If we let K = f −1(B) then we see that G/K is topologically isomorphic to Tn × E. Putting K = K ∩ N , we have that K is compact and f (K) = B. To see that K is a subgroup of G, let x and y be in K. Then x − y ∈ K, so there is a z ∈ K such that f (z) = f (x − y). This implies that x − y − z ∈ D, and since 3N ∩ D = {0} it follows that x − y − z = 0; that is, x − y ∈ K and so K is a subgroup of G. We claim that K = K + D. For if k ∈ K, there is a k ∈ K such that f (k) = f (k) and so k − k ∈ D. Thus K = K + D. By Exercises A5.4 #7, K is topologically isomorphic to K × D. Hence if θ is the canonical map of G onto G/K then θ(D) is topologically isomorphic to D and (G/K)/θ(D) is topologically isomorphic to G/K which is in turn topologically isomorphic to Tn × E. As θ(D) and E are discrete, Exercises A5.5 #7 tells us that G/K is locally isomorphic to Tn and hence also to Rn. Theorem A5.5.25 then says that G/K is topologically isomorphic to Ra × Tc × S, where S is a discrete group and a 0 and c 0. As G is compactly generated G/K and hence also S are compactly generated. So S is a discrete ﬁnitely generated abelian group and thus is topologically isomorphic to Zb × F , for some ﬁnite discrete group G and b 0. Exercises A5.12 587 1. (i) Let f be a continuous homomorphism of an LCA-group A into an LCA-group If f (A) is dense in B, show that the map f ∗ : B∗ → A∗, described in B. Proposition A5.9.5, is one-one. (ii) Show that if G is a compact Hausdorﬀ abelian group which is monothetic then G∗ is topologically iso
morphic to a subgroup of Td, the circle group endowed with the discrete topology. [Hint: Use (i) with A = Z and B = G.] (iii) Let A be an LCA-group which satisﬁes the duality theorem and B an LCAgroup. If f is a continuous one-one homomorphism of A into B show that f ∗(B∗) is dense in A∗. [Hint: See the proof of Corollary A5.11.4 and use Proposition A5.9.10 and Theorem A5.12.6.] (iv) Show that if G is a compact Hausdorﬀ abelian group with G∗ topologically isomorphic to a subgroup of Td, then G is monothetic. 2. Let A and B be LCA-groups and H a (not necessarily closed) ﬁnitely generated subgroup of A. If f is a continuous homomorphism of A into B such that the kernel of f lies wholly in H and is topologically isomorphic to Zn, for some n 1, and such that f (H) contains a subgroup topologically isomorphic to Z, show that H contains a subgroup topologically isomorphic to Zn+1. [Hint: Use Corollary A5.4.3.] 3. If H is a closed subgroup of an LCA-group G and g is an element of G not in H, show that there is a character γ of G such that γ(g) = 0 but γ(h) = 0, for all h ∈ H. 588 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 4. Let G be a locally compact Hausdorﬀ group. (i) Prove that G is a kω-space if and only if it is σ-compact. (See Exercises 10.3 #5 for the deﬁnition of kω-space.) (ii) If G is σ-compact, show that the Xn in the kω-decomposition can be chosen to be neighbourhoods of e. [Hint: As G is σ-compact, G = ∞ n=1 Yn where each Yn is compact. Let V be a compact symmetric neighbourhood of e and put Xn = Y1V ∪Y2V ∪· · ·∪YnV .] (iii) Every connected locally compact Hausdorﬀ group is a kω-space. (iv) G has an open neighbourhood U of 1 such that U is a compact neighbourhood of 1. (v) The subgroup gp (U ) of G generated algebraically by U is open in G. (vi) The subgroup gp (U ) of G generated algebraically by U is open in G, and so gp (U ) is an open locally compact σ-compact subgroup of G. (vii) (Glöckner et al. [160]) A topological space (X, τ ) is said to be a locally kω-space if each point in (X, τ ) has an open neighbourhood which is a kω-space. Prove that every locally compact Hausdorﬀ group is a locally kω-space. (viii) Verify that every discrete space, every compact Hausdorﬀ space, every kωspace, and every closed subspace of a locally kω-space is a locally kω-space. (ix) Verify that every metrizable kω-space is separable but not every metrizable locally kω-space is separable. (x) Prove that every locally kω-space is a k-space. 589 5. Let P be a property of topological groups; that is if topological groups G and H are topologically isomorphic then G has property P if and only if H has property P. The property P is said to be a three space property if whenever G is any topological group with a closed normal subgroup N and the topological groups N and G/N have property P, then G has property P. (See Bruguera and Tkachenko [64].) (i) Prove that “being a ﬁnite group” is a three space property. (ii) Prove that “being a ﬁnitely generated group” is a three space property. (iii) Prove that for any given inﬁnite cardinal m, “being a topological group of cardinality m” is a three space property. (iv) Prove that “being a discrete group” is a three space property. (v) Let G be a locally compact Hausdorﬀ group and N a closed normal subgroup of G. If f : G → G/N is the canonical map, show that for each compact subset C of G/N there exists a compact subset S of G such that f (S) = C. (vi) Deduce that if N is a closed normal subgroup of a locally compact Hausdorﬀ group G such that both N and G/N are compactly generated, then G is also compactly generated. So “being a compactly generated locally compact Hausdorﬀ group” is a three space property. (vii) If in (v), N is also compact show that f −1(C) is compact. (viii) Deduce that if G is a Hausdorﬀ topological group having a normal subgroup K such that both K and G/K are compact, then G is compact. So “being a compact Hausdorﬀ group” is a three space property. 590 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.13 The Duality Theorem and the Principal Structure Theorem Let A, B, and C be topological groups, f1 a A5.13.1 Deﬁnition. continuous homomorphism of A into B and f2 a continuous homomorphism of B into C. The sequence 0 −−→ A −−→ B −−→ C −−→ 0 f1 f2 is said to be exact if (i) f1 is one-one; (ii) f2 is onto; and (iii) the kernel of f2 equals f1(A). A5.13.2 Proposition. Let K be a compact subgroup of an LCA-group G, so that we have an exact sequence 591 0 −−→ K −−→ G −−→ G/C −−→ 0 f1 f2 where f2 is an open continuous homomorphism and f1 is a homeomorphism of K onto its image in G. Then the sequence f ∗ 2 f ∗ 1 is exact and f ∗ 0 ←−− K∗ ←−− G∗ ←−− (G/K)∗ ←−− 0 1 and f ∗ 2 are open continuous homomorphisms. By Proposition A5.9.5, f ∗ Proof. with Theorem A5.12.6 we see that f ∗ of f ∗ 2 equals the kernel of f ∗ 1 consider the diagram 2 is one-one. Using Corollary A5.11.5 together 1 is both open and onto. To see that the image 0 .......................................................................................................................................... ............ K G/K .................................................................................................................... ............ 0 G f1 .......................................................................................................................................... ............ f2 .................................................................................................................... ............ ................................................................................................................................................................................................................................................................................................................................................... ............ ............................................................................................................................................................................................................. ............ 1 f ∗ f ∗ 2 (γ) f ∗ 2 (γ) .......................................................................................................................................... γ ............ T Let γ be any character of G/K and k any element of K. Then 1 f ∗ f ∗ 2 γ(k) = γf2f1(k) = 0 1 f ∗ as the given sequence is exact. Therefore f ∗ 2 ⊆ Kernel f ∗ 1 . Now if φ ∈ G∗ and f ∗ 1 (φ) = 0, then we have φf1(k) = 0 for all k ∈ K. So there exists a homomorphism δ : G/K → T such that δf2 = φ. As f2 is both open and onto, δ is continuous. So Kernel f ∗ 2 . Hence Image f ∗ 2 = Kernel f ∗ 1 . 2 (γ) = 0 and so Image f ∗ 1 ⊆ Image f ∗ Finally we have to show that f ∗ 2 is an open map. Let C be a compact subset of G/K, U an open subset of T and P (C, U ) the set of all elements of (G/K)∗ which map C into U . Then P (C, U ) is a sub-basic open set in (G/K)∗. Now by Exercises A5.12 #5(i) there exists a compact subset S of G such that f2(S) = C. Thus we see that P (S, U ) is a sub-basic open subset of G∗ such that 2 is a homeomorphism of (G/K)∗ onto f ∗ 2 ((G/K)∗). So f ∗ 2 (P (C, U )) = P (S, U ) ∩ f ∗ 1 is open in G∗; that is, Image f ∗ its image in G∗. As K∗ is discrete, Kernel f ∗ 2 is open in G∗. So f ∗ 2 is an open map. 592 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.13.3 Proposition. Let A be an open subgroup of an LCA-group G, so that we have an exact sequence 0 −−→ A −−→ G −−→ G/A −−→ 0 f1 f2 where the homomorphisms f1 and f2 are open continuous maps. Then the sequence 0 ←−− A∗ ←−− G∗ ←−− (G/A)∗ ←−− 0 f ∗ 1 f ∗ 2 is exact, f ∗ its image in G∗. 1 is open and continuous and f ∗ 2 is a homeomorphism of (G/A)∗ onto By Proposition A5.9.5, f ∗ Proof. 2 = Kernel f ∗ f ∗ G/A is discrete and (G/A)∗ is compact. As f ∗ 2 is a homeomorphism of (G/A)∗ onto its image in G∗. f ∗ 2 is one-one. That Image 1 is proved exactly as in Proposition A5.13.2. As A is open in G, 2 is one-one and (G/A)∗ is compact, 1 is onto and f ∗ Finally we have to show that f ∗ 1 is an open map. Let K be a compact neighbourhood of 0 in G which lies in A. If Va is as in Corollary A5.7.7, then P (K, Va) is an open set in G∗ such that P (K, Va) is compact. Of course, f ∗ 1 (P (K, Va)) consists of those elements of A∗ which map K into Va, and so is open in A∗. If we put H equal to the group generated by f ∗ 1 (P (K, Va)) then H is an open subgroup of A∗. Furthermore as gp {P (K, Va))} is an open and closed subgroup of G∗, P (K, Va) ⊆ gp {P (K, Va)} = B. As B is generated by P (K, Va) it is σ-compact. The Open Mapping Theorem A5.4.4 then implies that f ∗ 1 : B → H is open. As B is an open subgroup of G∗ and H is an open subgroup of A∗, f ∗ 1 : G∗ → A∗ is open. The next Proposition is a corollary of the 5-Lemma of category theory. (See https://en.wikipedia.org/wiki/Short_five_lemma.) It is easily veriﬁed by “diagram- chasing". Let A, B, C, D, E and F be abelian topological A5.13.4 Proposition. groups and f1, f2, f3, f4, f5, f6 and f7 be continuous homomorphisms as indicated in the diagram below. 0 −−→ A f1 −−→ B f2 −−→ C −−→ 0 593    0 −−→ D −−→ E    f5 f3 f6 f7    −−→ F f4 −−→ 0 Let each of the horizontal sequences be exact and let the diagram be commutative (that is, f3f5 = f6f1 and f4f6 = f7f2). If f5 and f7 are algebraically isomorphisms (that is, both one-one and onto) then f6 is also an algebraic isomorphism. We now prove the Pontryagin van-Kampen Duality Theorem for compactly generated LCA-groups. 594 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be a compactly generated LCA-group and Γ A5.13.5 Theorem. its dual group. Then the canonical map α of G into Γ∗ is a topological group isomorphism of G onto Γ∗. By Proposition A5.12.10, G has a compact subgroup K such that G/K Proof. is topologically isomorphic to Ra × Zb × Tc × F , where F is a ﬁnite discrete abelian group and a, b, and c are non-negative integers. So we have an exact sequence 0 −
−→ K −−→ G −−→ G/K −−→ 0 f1 f2 Applying Proposition A5.13.2 to this sequence and Proposition A5.13.3 to the dual sequence, we obtain that the sequence 0 −−→ K∗∗ −−→ Γ∗ −−→ (G/K)∗∗ −−→ 0 f ∗∗ 1 f ∗∗ 2 is also exact. It is easily veriﬁed that the diagram 0 −−→ K    0 −−→ K∗∗ αK f1 −−→ G −−→ G/K f2 −−→ 0    α    αG/K −−→ Γ∗ −−→ (G/K)∗∗ −−→ 0 f ∗∗ 1 f ∗∗ 2 is commutative, where αK and αG/K are the canonical maps. As we have already seen that K and G/K satisfy the duality theorem, αK and αG/K are isomorphisms. This implies, by Proposition A5.13.4, that α is both topological an algebraic isomorphism. As α is continuous and G is compactly generated, the Open Mapping Theorem A5.4.4 then implies that α is an open map, and hence a topological isomorphism. At long last we can prove the Pontryagin van-Kampen Duality for all LCA-groups. 595 [Pontryagin-van Kampen Duality Theorem] Let G A5.13.6 Theorem. be an LCA-group and Γ its dual group. Then the canonical map α of G into Γ∗ is a topological group isomorphism of G onto Γ∗. By Corollary A5.12.8, G has an open subgroup A which is compactly generated. So we have an exact sequence 0 −−→ A −−→ G −−→ G/A −−→ 0 f1 f2 Applying Proposition A5.13.3 and then Proposition A5.13.2 yields the exact sequence 0 −−→ A∗∗ −−→ Γ∗ −−→ (G/A)∗∗ −−→ 0 f ∗∗ 1 f ∗∗ 2 and the commutative diagram 0 −−→ A f1 −−→ G −−→ G/A f2 −−→ 0    αA    α    αG/A 0 −−→ A∗∗ −−→ Γ∗ −−→ (G/A)∗∗ −−→ 0. f ∗∗ 1 f ∗∗ 2 As A is a compactly generated LCA-group and G/A is a discrete group, both A and G/A satisfy the duality theorem and so αA and αG/A are topological isomorphisms. By Proposition A5.13.4, α is an algebraic isomorphism. Since f1, f ∗∗ 1 and αA are all open maps and αf1 = f ∗∗ 1 αA, we see that α is also an open map, and hence a topological isomorphism. We can now prove the structure theorem for compactly generated LCA- groups, from which the Principal Structure Theorem for all LCA-groups is a trivial consequence. 596 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be a compactly generated LCA-group. Then G A5.13.7 Theorem. is topologically isomorphic to Ra × Zb × K, for some compact abelian group K and non-negative integers a and b. Proof. By Proposition A5.12.10, we have an exact sequence f1 f2 0 −−→ C −−→ G −−→ Ra × Zb × Tc × F −−→ 0 where C is a compact group, F is a ﬁnite discrete group and a, b and c are nonnegative integers. By Proposition A5.13.2, we, therefore, have an exact sequence f ∗ 2 f ∗ 1 0 ←−− C∗ ←−− G∗ ←−− Ra × Tb × Zc × F ←−− 0 2 is an open map. So G∗ has an open subgroup topologically isomorphic where f ∗ to Ra × Tb. As R and T are divisible groups, Proposition A5.3.8 says that G∗ is topologically isomorphic to Ra×Tb×D, for some discrete group D. As G satisﬁes the duality theorem, G is topologically isomorphic to G∗∗ which in turn is topologically isomorphic to Ra × Zb × K, where K is the compact group D∗. Since every LCA-group has an open compactly generated subgroup we obtain the Principal Structure Theorem for LCA-groups. A5.13.8 Theorem. [Principal Structure Theorem for LCA-Groups] Every LCA-group has an open subgroup topologically isomorphic to Ra × K, for some compact abelian group K and non-negative integer a. As an immediate consequence we have the following signiﬁcant result. A5.13.9 Theorem. topologically isomorphic to Ra × K, where K is a compact connected abelian group and a is a non-negative integer. Every connected LCA-group is 597 A5.13.10 Remarks. (i) Theorem A5.13.7 generalizes the well-known result that every ﬁnitely generated abelian group is the direct product of a ﬁnite number of copies of the inﬁnite cyclic group with a ﬁnite abelian group. (ii) One might suspect that one could improve upon the Principal Structure Theorem for LCA-Groups A5.13.8 and show that every LCA-group is topologically isomorphic to Ra × K × D, where K is compact, D is discrete and a is a nonnegative integer. Unfortunately, as the following example shows, this statement is false. ∞ i=1 ∞ i=1 A5.13.11 Example. Let G be the group Hi, where each Hi is a cyclic group of order four. Let C be the subgroup of G consisting of all elements g ∈ G such that 2g = 0. Then C is algebraically isomorphic to Ci, where each Ci is a cyclic group of order two. Put the discrete topology on each Ci and the product topology on C. So C is a compact totally disconnected topological group. Deﬁne a topology on G as follows: A base of open neighbourhoods at 0 in G consists of all the open subsets of C containing 0. With this topology G is a totally disconnected LCA-group having C as an open subgroup. Suppose that G is topologically isomorphic to Ra×K ×D, where K is a compact abelian group, D is a discrete abelian group, and a is a non-negative integer. As G is totally disconnected, a = 0. Further, as G is not compact, D must be inﬁnite. By the Principal Structure Theorem for LCA-Groups A5.13.8, G has an open subgroup H. As G is totally disconnected a = 0. topologically isomorphic to Ra ×C, where C is compact and a is a non-negative integer. Suppose that G is topologically isomorphic to H × D, where D is a discrete subgroup of G. As G is not compact D must be inﬁnite As D is discrete and C is compact, D ∩ C is ﬁnite. If x ∈ D, then 2x ∈ D ∩ C. If 2x = 2y, for y ∈ D, then 2(x − y) = 0, and so x − y ∈ D ∩ C. As D ∩ C is ﬁnite, 2x = 2y for only a ﬁnite number of y ∈ D. Hence D is ﬁnite. This is a contradiction and so our supposition is false. 598 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Exercises A5.13 1. Show that if G is an LCA-group such that G and its dual group are connected, then G is topologically isomorphic to Rn, for some non-negative integer n. 2. Show that an LCA-group G has enough continuous homomorphisms into R to separate points if and only if G is topologically isomorphic to Rn × D, where D is a discrete torsion-free abelian group. [Hint: a compact group admits no nontrivial continuous homomorphisms into R.] 3. Describe the compactly generated LCA-groups which are topologically isomorphic to their dual groups. 4. A topological group G is said to be solenoidal if there exists a continuous homomorphism f of R into G such that f (R) = G. (i) Show that if G is also locally compact Hausdorﬀ, then G is either a compact connected abelian group or is topologically isomorphic to R. [Hint : Observe that f (R) is topologically isomorphic to R1 × R2 × · · · × Rn × K, where each Ri is a copy of R. Let pi be the projection of f (R) onto Ri and note that pif is a continuous homomorphism of R into Ri.] (ii) Show that if G is a compact Hausdorﬀ solenoidal group then the dual group of G is topologically isomorphic to a subgroup of Rd, the additive group of real numbers with the discrete topology. 5. Let F be a ﬁeld with a topology such that the algebraic operations are (The additive structure of F , then, is an abelian topological continuous. group.) Show that if F is locally compact Hausdorﬀ and connected then F , as a topological group, is isomorphic to Rn, for some positive integer n. (A further analysis would show that F is either the real number ﬁeld R (n = 1), the complex number ﬁeld (n = 2) or the quaternionic ﬁeld (n = 4).) 599 6. Show that if G is any LCA-group then there exists a continuous one-one homomorphism β of G into a dense subgroup of a compact Hausdorﬀ abelian group. Prove this by two diﬀerent methods. [Hint: (1) Use the fact that any LCA-group has enough characters to separate points. (2) Alternatively, let Γ be the dual group of G and Γd the group Γ endowed with the discrete topology. Put K = (Γd)∗ and let β be deﬁned by (g, γ) = (γ, β(g)), g ∈ G, γ ∈ Γ. The group K = (Γd)∗ is called the Bohr compactiﬁcation of G. Harald Bohr (left) and Niels Bohr Harald Bohr (1887–1951) was a Jewish Danish mathematician and younger brother of the Nobel prize-winning physicist Niels Bohr. Harald did research in mathematical analysis with the Cambridge University mathematician, G.H. Hardy, and Harald was the founder of the theory of almost-periodic functions. He was also a member of the Danish Football Team which won a silver medal in the 1908 Olympics.] 7. A topological group G is said to be maximally almost periodic (MAP) if there exists a continuous one-to-one homomorphism of G into a compact Hausdorﬀ group. (i) Verify that every compact Hausdorﬀ is maximally almost periodic. (ii) Using Exercises A5.13 #6, verify that every LCA-group is a MAP-group. 600 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 8. A topological group G is said to be minimally almost periodic (m.a.p.) (von Neumann [411], von Neumann and Wigner [412]) if there every continuous homomorphism φ of G into every compact Hausdorﬀ group satisﬁes φ(g) = 1, for each g ∈ G; that is, every continuous homomorphism of G into a compact Hausdorﬀ group is trivial. A group H is said to be simple if it has no normal subgroups other than 1 and H. A topological group S is said to be topologically simple if its only closed normal subgroups are 1 and S. (i) Verify that every simple group with the discrete topology is topologically simple. (ii) Verify that every inﬁnite topologically simple group is minimally almost periodic. (iii) Verify that no inﬁnite abelian group is a simple group. It is easy to verify that there exist inﬁnite simple groups, which by (i) and (ii) above show that there exist minimally almost periodic groups. Let ℵ be any inﬁnite cardinal number and S a set of cardinality ℵ. Let Aℵ be the group of all even permutations of S which ﬁx all but a ﬁnite number of members of S. That Aℵ is a simple group follows easily from 3.2.4 of Robinson [342] The concept of minimally periodic group was introduced by John von Neumann and in a subsequent paper with Nobel prize-winning physicist Eugene Paul Wigner (1902-1995) proved that the Special Linear group SL(2, C) with even the discrete topology is a m.a.p. group. Dieter Remus (Remus [338]) proved that every free abelian group and every inﬁnite divisible abelian group admits a m.a.p. to
pology. Saak Gabriyeylan (Gabriyelyan [146]) proved that every inﬁnitely-generated abelian group admits a m.a.p. topology. Wigner 601 9. Let Γ be any LCA-group and γ1, . . . , γn ∈ Γ. If φ is any homomorphism of Γ into T, show that there is a continuous homomorphism ψ of Γ into T such that \|ψ(γi) − φ(γi)\| < ε, i = 1, . . . , n. [Hint: Use Exercises A5.13 #6, method (2).] 10. A topological group G is said to be almost connected if G/G0 is a compact group, where G0 is the connected component of the identity in G. (i) Using Exercises A5.13 #5(vi), prove that every almost connected locally compact group is compactly generated. (ii) Deduce from (i) above and Theorem A5.13.7 that every almost connected locally compact abelian group is topologically isomorphic to Rn × K, where K is a compact abelian group and n is a non-negative integer. (iii) Using Exercises A5.12.5(vi) and (ii) above, prove that “being an almost connected LCA-group” is a three space property. (See Exrercises A5.12 #5 for the deﬁnition and examples of the three space property”.) A5.14 Consequences of the Duality Theorem Part One: Kronecker’s Theorem We begin this section by showing that the dual of a subgroup is a quotient group and the dual of a quotient group is a subgroup. Let H be a closed subgroup of the LCA-group G and A5.14.1 Deﬁnition. Ω the set of all γ in the dual group Γ of G such that (h, γ) = 0, for all h ∈ H. Then Ω is called the annihilator of H. For ﬁxed h ∈ H, the continuity of (h, γ) shows that the set of all γ with (h, γ) = 0 is closed, so that Ω is the intersection of closed sets and is therefore closed. Clearly Ω is a group and so it is a closed subgroup of Γ. 602 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE A5.14.2 Proposition. With the above notation, if Ω is the annihilator of H, then H is the annihilator of Ω. Proof. Corollary A5.12.7 there is a γ ∈ Ω such that (g, γ) = 0. If h ∈ H, then (h, γ) = 0 for all γ ∈ Ω. If g ∈ G and g ∈ H then, by A5.14.3 Theorem. If Γ is the dual group of G and Ω is the annihilator of H, then Ω and Γ/Ω are topologically isomorphic to the dual groups of G/H and H, respectively. Let H be a closed subgroup of an LCA-group G. Let f be the canonical homomorphism of G onto G/H. Then Proof. Proposition A5.9.5 says that the map f ∗ : (G/H)∗ → G∗ is a continuous one-one homomorphism. Further, the last paragraph of the proof of Proposition A5.13.3 shows that f ∗ is a homeomorphism of (G/H)∗ onto its image in G∗. Of course the deﬁnition of f ∗ tells us that f ∗((G/H)∗) = Ω, and so Ω is topologically isomorphic to (G/H)∗. The fact that Γ/Ω is topologically isomorphic to H∗ then follows from the above, together with Proposition A5.14.2 and the duality theorem. As a corollary we have the following generalization of Corollary A5.11.4. A5.14.4 Corollary. every character on H can be extended to a character on G. If H is a closed subgroup of an LCA-group G, then If φ is a character on H, then by Theorem A5.14.3, φ ∈ Γ/Ω. If f is the Proof. canonical homomorphism of Γ onto Γ/Ω and f (γ) = φ, γ ∈ Γ, then (h, γ) = (h, φ), for all h ∈ H. So γ is the required extension of φ. We now record an application in the area of diophantine approximation. Firstly observe that the deﬁnition of the annihilator of a subgroup H of an LCA-group G would make sense even if H were not closed in G. However it is obvious that the annihilator of H, Ω(H), would equal Ω(H). We will see that this observation is quite useful. 603 Let G be a (not necessarily closed) subgroup of Rn, A5.14.5 Proposition. n 1. Let Ω(G) denote the annihilator of G in (Rn)∗ = Rn. then Ω(G) = {y ∈ Rn : (y\|x) is an integer for each x ∈ G}, where y = (y1, . . . , yn) ∈ Rn, x = (x1, . . . , xn) ∈ G ⊆ Rn and (y\|x) = n i=1 yixi. Proof. Exercise. As an immediate consequence of Proposition A5.14.2 we have the following result. A5.14.6 Proposition. G. If G is any subgroup of Rn, n 1, then Ω(Ω(G)) = Translating Proposition A5.14.6 using Proposition A5.14.5 we obtain a characterization of those points lying in the closure of a subgroup of Rn. A point x lies in the closure of a subgroup G of A5.14.7 Proposition. Rn, n 1 if and only if (y\|x) is an integer for all y ∈ Rn such that (y\|g) is an integer for all g ∈ G. 604 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE We apply this characterization to the case where G is the subgroup of Rn, n 1 generated by the vectors e1, . . . , en of the canonical basis and by an arbitrary number m of points ai, i = 1, . . . , m of Rn. To say that (y\|ei) is an integer, for i = 1, . . . , n means that each coordinate of y is an integer. So we obtain the following theorem of the German mathematician Leopold Kronecker (1823–1891). [Kronecker once said: “Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk” (“God made the integers, all else is the work of man.”)] Kronecker (Kronecker) Let ai = (ai1, . . . , ain) for i = 1, . . . , m A5.14.8 Theorem. and b = (b1, . . . , bn) be points of Rn, n 1. In order that for each ε > 0 there exists integers q1, . . . , qm and integers p1, . . . , pn such that q1a1j + q2a2j + · · · + qmamj − pjbj ε for j = 1, . . . , n it is necessary and suﬃcient that for each ﬁnite sequence r1, . . . , rn of integers such that the numbers n j=1 aijrj, for i = 1, . . . , m, are all integers, the number n j=1 bjrj should also be an integer. Putting m = 1 we obtain the following corollary which is a generalization of Exercise Set A5.5 #3 (ii). 605 Let θ1, . . . , θn be real numbers. A5.4.19 Corollary. In order that, given any n real numbers x1, . . . , xn and a real number ε > 0, there should exist an integer q and n integers pj such that ε for j = 1, . . . , n qθj − pj − xj it is necessary and suﬃcient that there exist no relation of the form n j=1 rjθj = h , where the rj are integers not all zero and h is an integer. (In particular this implies that the θj and the ratios θj/θk, j = k, must be irrational.) For some further comments in this area of approximation see Hewitt and Ross [180, §26.19] and Bourbaki [51, §1.3 of Vol. 2]. Exercises A5.14 1. Prove that a closed subgroup of a compactly generated LCA-group is compactly generated. Kenneth Ross in Ross [346] deﬁned a (necessarily compactly generated) topological group a Moskowitz-Morris group if each closed subgroup of it is compactly generated. This exercise shows that every compactly generated LCA-group is a Moskowitz-Morris group. [Hint: Use Corollary A5.5.18.] 2. (i) Show that any closed subgroup of Rn × Tm is topologically isomorphic to Ra × Zb × Tc × D, where D is a discrete ﬁnite group, a + b n and c m. (ii) Show that any Hausdorﬀ quotient group of Rn × Tm is topologically isomorphic to Ra × Tb with a n and a + b n + m. 606 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE (iii) Show that any closed subgroup of Rn × Tm × F , where F is a discrete free abelian group, is topologically isomorphic to Ra × Zb × Tc × D × F , where a + b n, c m, D is a discrete ﬁnite group and F is a subgroup of F . (iv) Show that any closed subgroup of Rn × Tm × D, where D is a discrete abelian group, is topologically isomorphic to Ra × Tb × D, where D is a discrete group, a n and b m. (v) Show that any Hausdorﬀ quotient of Rn × Tm × D, where D is a discrete abelian group, is topologically isomorphic to Ra × Tb × D, where D is a discrete group, a n and a + b n + m. (vi) Hence show that any closed subgroup or Hausdorﬀ quotient group of Rn×Zm×K, where K is a compact Hausdorﬀ abelian group, is topologically isomorphic to Ra × Zb × K, where K is a compact group, a n and a + b n + m. 3. Let G be a compactly generated LCA-group. Assuming the structural result Proposition A5.12.5 which says that G has a subgroup A topologically isomorphic to Zn, n 0, such that G/A is compact, show that G∗ is locally isomorphic to Rn. Hence prove that G is topologically isomorphic to Ra × Zb × K, where a + b = n and K is compact. 4. Let G be any LCA-group. Assuming the structural result Theorem A5.13.9 which says that every connected LCA-group is topologically isomorphic to Ra × K, for some compact group K and a 0, prove the Principal Structure Theorem A5.13.9. [Hint: Let C be the component of 0 in G and observe that Exercises A5.3 #4(i) says that G/C has a compact open subgroup A. So G has an open subgroup H with the property that H/C is compact. Deduce that H∗ is locally isomorphic to Rn.] 5. Let G be a compactly generated LCA-group. Assuming the Principal Structure Theorem A5.13.9 show that G is topologically isomorphic to Ra ×Zb ×K, where K is compact, a 0 and b 0. [Hint: Observe that G∗ has an open subgroup topologically isomorphic to Ra × K1, where K1 is compact and a 0. Noting that K∗ 1 is a quotient 607 group of G, we see that it must be ﬁnitely generated and so K1 is topologically isomorphic to Tb × F , where F is a ﬁnite discrete group and b 0.] 6. Let G be a subgroup of Rn, n 1 and Ω(G) its annihilator. Show that Ω(G) = {y ∈ Rn : (y\|x) is an integer for each x ∈ G}, where y = (y1, . . . , yn) ∈ Rn, x = (x1, . . . , xn) ∈ G ⊆ Rn and (y\|x) = n [Hint: Recall that T = R/Z and use Examples A5.7.3 (ii) and Theorem A5.9.1.] n=1 yixi. 7. Recall that a topological group G is said to be almost connected if the quotient group G/G0 is compact, where G0 is the connected component of the identity If G is an almost connected LCA-group, show that G is element of G. topologically isomorphic to Rn × K, for some compact group K and n 0. A5.15 Consequences of the Duality Theorem Part Two: Connected and Totally Disconnected LCA-groups A5.15.1 Theorem. G is metrizable if and only if Γ is σ-compact. Let G be an LCA-group and Γ its dual group. Then Assume that G is metrizable. Then G has a countable base of compact Proof. neighbourhoods U1, U2, . . . , Un, . . . of 0. By Corollary A5.7.7, if a = 1 5 , then the sets P (Ui, Va), i = 1, 2, . . . , are compact neighbourhoods of 0 in Γ. As each γ ∈ Γ is continuous, Γ = P (Ui, Va). So Γ is σ-compact. ∞ i=1 Conversely, assume that Γ is σ-compact. By Exercise
s A5.12 #4, there exists a family {Yn}, n = 1, 2, . . . of compact neighbourhoods of 0 in Γ such that every compact subset of Γ lies in some Yn and Yn ⊆ Yn+1, n 1. So the family {g ∈ G : (g, γ) ∈ V1/k, for all γ ∈ Yn}, for k = 2, 3, . . . and n = 1, 2, . . . is a base of neighbourhoods of 0 in G. (Observe that in saying that these sets are neighbourhoods we are using the fact that G is the dual group of Γ.) Thus G has a countable base of neighbourhoods at 0 which, by Exercises A5.6 #3(ii), implies that G is metrizable. 608 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE As a corollary we have the following striking result. A5.15.2 Corollary. metrizable if and only if its dual group Γ is countable. Let G be an LCA-group. Then G is compact and Proof. Exercise. We now characterize those compact groups which are connected. Let G be a locally compact Hausdorﬀ group and A5.15.3 Proposition. K a compact subset of G. Then there exists an open and closed compactly generated subgroup of G containing K. Proof. Exercise. An element g of a topological group G is said to be A5.5.4 Deﬁnition. compact if gp {g}, the smallest closed subgroup of G containing g, is compact. A5.15.5 Example. and a 0 and b 0, is compact if and only if it lies in {0} × {0} × K. An element g ∈ G = Ra × Zb × K, where K is compact A5.15.6 Proposition. elements is a closed subgroup of G. If G is an LCA-group, then the set S of compact 609 If g and h are in S, then g − h ∈ gp {g} + gp {h} ⊆ gp {g} + gp {h}. As Proof. g − h lies in the compact group gp {g} + gp {h}, it is compact and so is in S. Thus S is a subgroup of G. Now let x ∈ S. By Proposition A5.15.3 there is an open compactly generated subgroup H of G containing x. By Theorem A5.13.7, H is topologically isomorphic to Ra × Zb × K, where K is compact, a 0 and b 0. To see that x ∈ S we have to show only that the Ra coordinate of x and the Zb coordinate of x are both zero. If the Ra coordinate of x or the Zb coordinate of x were diﬀerent from zero, then there would be an entire neighbourhood of x disjoint from S–since all compact elements of Ra × Zb × K lie in {0} × {0} × K. This would contradict the fact that x ∈ S. A5.15.7 Remark. The set of compact elements of a non-abelian locally compact Hausdorﬀ group G need not be a subgroup of G. As an example of this, let G be the discrete group generated by two elements a and b with the relations a2 = b2 = e. Then a and b are compact elements, but ab is not compact. 610 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be an LCA-group, Γ its dual group, S the set A5.15.8 Theorem. of compact elements in Γ and C the component of 0 in G. Then S is the annihilator in Γ of C and C is the annihilator in G of S. Proof. (i) Assume that there is a compact subgroup K of Γ such that K = {0}. Then G has a discrete quotient group K∗. As K∗ is not connected, G is not connected. (ii) Assume that G is totally disconnected. Let γ ∈ Γ and U be a neighbourhood of 0 in G such that γ(g) ∈ V1/4, for all g ∈ U . As G is locally compact and totally disconnected, Exercises A5.3 #4(i) implies that U contains a compact open subgroup K. It is clear that γ(K) is a subgroup of V1/4 and hence equals {0}; that is, γ ∈ Λ(Γ, K) the annihilator in Γ of K. As K is open in G, Λ(Γ, K), is compact. So every γ ∈ Γ lies in a compact group. Hence S = Γ. (iii) Now let G be any LCA-group. Then G/C is a totally disconnected LCA-group and so, by (ii), the dual group of G/C contains only compact elements. As (G/C)∗ is topologically isomorphic to Λ(Γ, C), we see that Λ(Γ, C) ⊆ S. As C is a connected LCA-group (i) implies that C∗ has no nontrivial compact subgroup; that is, Γ/Λ(Γ, C) has no compact subgroup. Suppose γ ∈ S and γ ∈ Λ(Γ, C). Then there is a compact subgroup K of Γ whose image under the canonical map f : Γ → Γ/Λ(Γ, C) is not {0}. This image is a compact subgroup of Γ/Λ(Γ, C) and so we have a contradiction. Hence S ⊆ Λ(Γ, C). Thus S = Λ(Γ, C). The dual statement then follows from Proposition A5.14.2. A5.15.9 Corollary. disconnected if and only if every element in G∗ is compact. Let G be an LCA-group. Then G is totally A5.15.10 Corollary. only if G∗ has no compact subgroup = {0}. Let G be an LCA-group. Then G is connected if and Let G be a compact Hausdorﬀ abelian group with C A5.15.11 Corollary. the component of 0 in G. Let Φ be the torsion subgroup of Γ (that is, the subgroup consisting of all elements of ﬁnite order). Then Φ = Λ(Γ, C), the annihilator in Γ of C, and C = Λ(G, Φ). Also Φ is isomorphic to (G/C)∗. 611 Proof. The ﬁrst two equalities follow from the Theorem A5.15.7, since an element of a discrete group is compact if and only if it has ﬁnite order. The last statement follows from the duality between subgroups and quotient groups. As an immediate consequence of Corollary A5.15.11 we have an interesting characterization of those compact groups which are connected. A5.15.12 Corollary. and only if its dual group is torsion free. A compact Hausdorﬀ abelian group is connected if Notation. Let G be an abelian group and fn the homomorphism of G into itself given by fn(g) = ng, for n a positive integer. We denote fn(G) by G(n) and f −1 n {0} by G(n). Let G be an LCA-group with dual group Γ. For A5.15.13 Proposition. any n ∈ N, (i) Λ(Γ, G(n)) = Γ(n), and (ii) Λ(Γ, G(n)) = Γ(n). Proof. and so γ(ng) = nγ(g) = 0. Thus γ ∈ Γ(n). Conversely, nγ(g) = γ(ng) = 0, for all g ∈ G, and so γ ∈ Λ(Γ, G(n)). Hence (i) is true. Let γ ∈ Λ(Γ, G(n)). Then for each g ∈ G we have ng ∈ G(n) if γ ∈ Γ(n), then To prove (ii) regard G as the dual group of Γ. Then, by (i), Λ(G, Γ(n)) = G(n) Γ(n) = Λ(Γ, Λ(G, Γ(n))) = Λ(Γ, Λ(G, Γ(n))) = Λ(Γ, G(n)). and so 612 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE Let G be an LCA-group and Γ its dual group. If G A5.15.14 Theorem. is divisible, then Γ is torsion-free. If Γ is torsion-free, then G(n) is dense in G, for n ∈ N. Let G be discrete or compact. Then G is divisible if and only if Γ is torsion-free. If G is divisible then G(n) = G, for all n and so Proposition A5.15.13(i) Proof. implies that Γ(n) = {0}; that is, Γ is torsion-free. If Γ is torsion-free then Proposition A5.15.13(ii) shows that Λ(G, Γ(n)) = G = G(n), for each n ∈ N. If G is discrete then G(n) is certainly closed, for each n ∈ N. If G is compact then G(n) is a continuous image of G and so is compact and closed. So in both cases G = G(n), for all n ∈ N, and G is divisible. A5.15.15 Corollary. Let G be a compact Hausdorﬀ abelian group and Γ its dual group. Then the following are equivalent: (i) G is connected; (ii) Γ is torsion-free; (iii) G is divisible. A5.15.16 Corollary. Every connected LCA-group G is divisible. Proof. G is topologically isomorphic to Rn × K, where K is compact and connected, from which the result immediately follows. A5.15.17 Remarks. It is also true that a compact Hausdorﬀ non-abelian group is connected if and only if it is divisible. (See Hofmann and Morris [191, Theorem 9.35] or Mycielski [299].) Corollary A5.15.16 does not extend to the non-abelian case. (See Hewitt and Ross [180, §24.44].) However, it is true that a connected locally compact group is somewhat divisible, where a group G is deﬁned to be somewhat divisible if it has subgroups H1, H2, . . . , Hn such that each Hi is a divisible group and G = H1.H2. . . . .Hn. (See Morris [294].) 613 1. Let G be an LCA-group. Show that it is compact and metrizable if and only if Exercises A5.15 its dual group is countable. [Hint: Use Corollary A5.4.3.] 2. Show that the Bohr-compactiﬁcation bG of an LCA-group G is metrizable if and only if G is compact and metrizable and bG = G. [Hint: See Exercises A5.13 #6 and use Exercises A5.15 #1 above.] 3. Let K be a compact subset of a locally compact Hausdorﬀ group G. Show that there exists an open and closed compactly generated subgroup of G containing K. 4. Let G be an LCA-group and Γ its dual group. Let a be the least cardinal number of an open basis at 0 in G and b be the least cardinal number of a family of compact subsets of Γ whose union is Γ. Show that a = b. [Hint: To prove a b, let S = {Ai : i ∈ I} be a family of compact sets such that Γ = Ai and b equals the cardinality of the index set I. Let Bi be an i∈I open set containing Ai such that Bi is compact, for each i ∈ I. Let S be the family of all ﬁnite unions of sets Bi1 ∪ Bi2 ∪ · · · ∪ Bin. Verify that Γ is the union of the members of S and that every compact subset of Γ is a subset of a member of S. Now proceed as in the proof of Theorem A5.15.1.] 5. (i) Let G be a compact Hausdorﬀ abelian group and w(G) the least cardinal number of an open basis of G. Show that w(G) equals the cardinal number of G∗. [Hint: Use Exercises A5.15 #4 above.] 614 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE (ii) If G is a compactly generated LCA-group, show that w(G) = w(G∗). (iii) If G is any LCA-group, show that w(G) = w(G∗). If G is ﬁnite, the result is trivial, so assume G is inﬁnite. Then G [Hint: has an open subgroup H topologically isomorphic to Rn × K, where K is compact and n 0. Show that w(G) = max [w(Rn × K), cardinal number of G/H] . Observing that G∗ has a compact subgroup A, topologically isomorphic to (G/H)∗ such that G∗/A is topologically isomorphic to Rn × K∗, show that w(G∗) max{w(Rn × K∗), w((G/H)∗).}] 6. Let G be a compact Hausdorﬀ abelian group with dual group Γ. Show that the following conditions are equivalent (where c denotes the cardinality of R). (i) G is solenoidal; (ii) Γ is algebraically isomorphic to a subgroup of R; (iii) Γ is torsion-free and the cardinal number of Γ c; (iv) G is connected and w(G) c. [Hint: See Exercises A5.13 #4. Assume the fact, from abelian group theory, that (iii) implies (ii).] 7. Let G be a divisible compact Hausdorﬀ (not necessarily abelian) group. Prove that G is connected. [Hint: Suppose that G is not connected and arrive at a contradiction by showing that G has a proper open normal subgroup H, such that G/H is a ﬁnite divisible group.] 8. Prove the following statements: (i) If group
s G1, G2, . . . , Gn are somewhat divisible, then the product group G1 × G2 · · · × Gn is somewhat divisible. (ii) Any quotient group of a somewhat divisible group is somewhat divisible. (iii) An abelian group is divisible if and only if it is somewhat divisible. 615 A5.99 Credit for Images 1. Leon Battista Alberti. Creative Commons License. https://tinyurl.com/yc9827oh 2. Harald and Niels Bohr. Public Domain. 3. Filippo Brunelleschi. Creative Commons License https://tinyurl.com/ya5qp4oh 4. Élie Cartan. Fair Use. https://en.wikipedia.org/wiki/File:Elie_Cartan.jpg 5. Rudolf Friedrich Alfred Clebsch. Public Domain. 6. Jean Alexandre Eugène Dieudonné. The Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO) owns the copyright and the photo can be used on the terms of the Creative Commons License Oberwolfach Photo Collection. Photo by Konrad Jacobs of Erlangen. https://opc.mfo.de/ Fair use. https: //en.wikipedia.org/wiki/File:JeanDieudonnØ.jpg 7. Per Enﬂo. The Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO) owns the copyright and the photo can be used on the terms of the GNU Free Documentation License. Photo by George M. Bergman. https://tinyurl.com/ y9pnasqm 8. Per Enﬂo, Stanisław Mazur, and goose prize. Public Domain. 9. Albert Einstein. Public Domain. 10. Israil Moiseevic Gelfand. The Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO) owns the copyright and the photo can be used on the terms of the Creative Commons License Oberwolfach Photo Collection. Photo by Konrad Jacobs of Erlangen. https://opc.mfo.de/ Photo by Konrad Jacobs, Erlangen. 11. Andrew Mattei Gleason. Copyright CC BY-SA 3.0 https://tinyurl.com/pqyhenr by https://tinyurl.com/yb7ahgyh Notices of the American Mathematical Society, November 2009. 12. Werner Heisenberg. Creative Commons Attribution-Share Alike Germany license. https://tinyurl.com/y7oyype7 13. David Hilbert. Public Domain. 14. Karl Hofmann & Sidney Morris at work. Photo owned by author. 616 APPENDIX 5: TOPOLOGICAL GROUPS: A GRADUATE COURSE 15. Hofmann and Morris August 2016. Photo owned by author. 16. Kenkichi Iwasawa. Public Domain. 17. Wilhelm Karl Joseph Killing. Public Domain. 18. Felix Klein. Public Domain. 19. Leopold Kronecker. Public Domain. 20. Richard Kenneth Lashof. Copyright George Bergman. Oberwolfach Photo Collection. GNU Free Documentation License. https://tinyurl.com/ya2bdnpr 21. (Marius) Sophus Lie. Public Domain. 22. Carl Louis Ferdinand von Lindemann. Public Domain. 23. Hendrik Antoon Lorentz. Public Domain. 24. Stanisław Mazur. Public Domain. 25. Deane Montgomery. Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO). Photo by Konrad Jacobs of Erlangen. 26. Julius Plücker. Public Domain. 27. Pierre Laurent Wantzel. Public Domain. 28. Perspective photo. Creative Commons License. https://tinyurl.com/ycs5g7qh 29. Lev Semyonovich Pontryagin. Creative Commons Attribution-ShareAlike 3.0 License. Monument to Lev Pontryagin on wall of building on Leninsky Prospekt in Moscow, where he lived from 1939 to 1988. Translation: In this home from 1939 to 1988 lived and worked the great mathematician Lev Semenovich Pontryagin. Author: Anatoly Terentiev. https://tinyurl.com/yb8qdajo 30. School of Athens. Public Domain. Copyuright 31. Władysław Hugo Dionizy Steinhaus. Public Domain. 32. Terence Chi-Shen Tao. GNU Free Documentation License. https://tinyurl. com/yc25cryk 33. Stanisław Marcin Ulam. Public Domain. 34. Egbert Rudolf van Kampen. Public Domain. 35. Hermann Weyl. ETH-Bibliothek ZÃĳrich, Bildarchiv. “This work is free and may be used by anyone for any purpose.” https://tinyurl.com/Hermann-Weyl 617 36. Eugene Paul Wigner. Public Domain. 37. Hidehiko Yamabe. Public Domain. 38. Leo Zippin. Copyright Mathematisches Forschungsinstitut Oberwolfach gGmbH (MFO) licensed with CC BY-SA 2.0 de https://tinyurl.com/pqyhenr. Photo by Konrad Jacobs. Appendix 6: Filters and Nets §A6.0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619 §A6.1 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 620 §A6.2 Filterbases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 640 §A6.3 Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 649 §A6.4 Wallman Compactiﬁcations: An Application of Ultraﬁlters . . . . . . . . . . . . . 659 618 Introduction36 619 In the theory of metric spaces, convergent sequences play a key role. This is because they can be used to capture everything about the topology of the metric space. For example, a set S in a metric space (X, d) is closed if and only if every convergent sequence of points in S converges to a point in S. So convergent sequences determine the closed sets, the open sets, the continuous functions and the topological properties. For example, a mapping between metric spaces is continuous if and only if it maps convergent sequences to convergent sequences, and a subset C of a metric space is compact if and only if every sequence consisting of points in C has a subsequence which converges. However, every metric space (X, d) satisﬁes the ﬁrst axion of countability, that is, every point in X has a countable base of neighbourhoods or, in other words, for each point a ∈ X there is a countable set of neighbourhoods U1, U2, . . . , Un, . . . such that every neighbourhood of a contains a Un for some n ∈ N . [We can choose Un = {x ∈ X : d(x, a) < 1 n}, for each n ∈ N.] However, a general topological space does not necessarily satisfy the ﬁrst axiom of countability; that is, it need not have a countable base of neighbourhoods for each point. For this reason convergent sequences, which are of course countable sets, do not capture the topology of a general toplogical space. Therefore we need a more general concept than convergent sequences which is rich enough to capture the topology of general topological spaces. In this appendix we introduce the notion of a ﬁlter. Filters do indeed capture everything about the topology of a general topological space. In particular, we shall see how closedness, continuity, and compactness can be expressed in terms of ﬁlters. We shall also see how ﬁlters can be used to give an alternative proof of the powerful Tychonoﬀ Theorem. There is an equivalent, but less elegant, generalization of convergent sequences which uses nets rather than ﬁlters. It is interesting to note that ﬁlters, which were introduced by Cartan [69, 70] in 1937, have been the preferred approach of many 36Two 14 minute YouTube videos provide a very gentle introduction to this Appendix, and in particular to §6.3. They are:- Topology Without Tears – Sequences and Nets – Video 3a – http://youtu.be/wXkNgyVgOJE & Video 3b –http://youtu.be/xNqLF8GsRFE . 620 APPENDIX 6: FILTERS AND NETS European (especially, French) mathematicians, while American mathematicians have tended to prefer nets, which were introduced in 1922 by E.H. Moore and H.L. Smith [289]. Bourbaki [51] sum up their philospophy on ﬂlters: ’it replaces to advantage the notion of “Moore-Smith convergence”’. Here we treat both ﬁlters and nets. A6.1 Filters §A6.1 introduces ﬁlters and ultraﬁlters on any non-empty set X. Filters and ultraﬁlters are then related to the notion of a topology on the set X. It is shown how the property of Hausdorﬀness can be expressed in terms of ﬁlters or ultraﬁlters. But the magic of ultraﬁlters, in particular, is demonstrated in an elegant proof of the powerful Tychonoﬀ Theorem for compact spaces. So we begin with the deﬁnition of a ﬂiter and ﬁnd examples. Then we introduce the notion of an ultraﬁlter and verify that for every ﬁlter there is a ﬁner ﬁlter which is an ultraﬁlter. We ﬁnd a beautiful characterization of ultraﬁlters amongst all ﬁlters. Let X be a set and F a set of subsets of X. Then F A6.1.1 Deﬁnition. is said to be a ﬁlter on X if (i) F1, F2 ∈ F implies F1 ∩ F2 ∈ F ; (ii) F ∈ F and F ⊆ G ⊆ X imply G ∈ F ; and (iii) Ø /∈ F . 621 A6.1.2 Remarks. (i) If F is a ﬁlter on a set X, then by Deﬁnition A6.1.1(ii), X ∈ F . (ii) If F is a ﬁlter on a set X, then it is not a topology on X. [This is clear since Ø /∈ F .] (iii) If τ is a topology on a set X, then τ is not a ﬁlter on X. [Again this is clear, since Ø ∈ τ .] (iv) If F is a ﬁlter on a set X, then τ = F ∪ {Ø} is a topology on X. [We see that X ∈ τ , Ø ∈ τ , and unions and ﬁnite interesections of sets in τ are in τ by Deﬁnition A6.1.1 (ii) and Deﬁnition A6.1.1(i), respectively. So τ is indeed a topology on X.] (v) If τ is a topology on a set X, then τ \ {Ø} is not necessarily a ﬁlter on X. [For example, if τ is the Euclidean topology on R, then S = τ \ {Ø} is not a ﬁlter since the open interval (0, 1) is a member of S, while the closed interval [0, 1] is not a member of S, and so S does not satisfy Deﬁnition A6.1.1(ii).] (vi) If F is a ﬁlter on a set X such that Fi∈F ﬁlter. Fi = Ø, then F is said to be a free A6.1.3 Examples. (i) Let X be any non-empty set and F consist of just the set X. Then F is a ﬁlter on X. (ii) Let X be any non-empty set and S a subset of X. If F consists of S and all subsets of X which contain S, then F is a ﬁlter on X and is called the principal ﬁlter generated by S (iii) Let X be an inﬁnite set and F = {F : Ø = F ⊆ X, X\F is a ﬁnite subset of X}. Then F is a ﬁlter on X and is called the Fréchet ﬁlter. Every Fréchet ﬁlter is a free ﬁlter and every free ﬁlter contains the Fréchet ﬁlter. [Verify this.] (iv) Let (X, τ ) be a topological space. For each x ∈ X, the set of all neighbourhoods of x is a ﬁlter. This ﬁlter is known as the neighbourhood ﬁlter, Nx, of the point x in (X, τ ). A neighbourhood ﬁlter is clearly not a free ﬁlter. (v) Let f be a mapping of a set X into a set Y and F a ﬁlter on X. Then f (F) is a ﬁlter on Y if and only if f is surjective. Further, f −1(f (F)) = F . [Verify these
statements.] 622 APPENDIX 6: FILTERS AND NETS A6.1.4 Proposition. (i) If F is a ﬂiter on a set X, then F has the ﬁnite intersection property; that is, if F1, F2, . . . , Fn ∈ F, n ∈ N, then F1 ∩ F2 ∩ · · · ∩ Fn = Ø; (ii) Let S be a set of subsets of a non-empty set X. There exists a ﬁlter F on X such that S ⊆ F if and only if S has the ﬁnite intersection property. Proof. Part (i) follows from (i) and (iii) of Deﬁntion A6.1.1 using mathematical induction. If F is a ﬁlter containing S, then it follows from (i) that F , and hence also S, has the ﬁnite intersection property. Conversely, if S has the ﬁnite intersection property, deﬁne F = {F : F ⊆ X such that there exist S1, S2, . . . , Sn ∈ S, n ∈ N with n i=1 Si ⊆ F.} Clearly F satisﬁes Deﬁnition A6.1.1 and S ⊆ F , and so part (ii) is proved. Using Proposition A6.1.4(ii) we introduce the following deﬁnition. A6.1.5 Deﬁnition. set X. If S has the ﬁnite intersection property, then the ﬁlter Let S be a non-empty set of subsets of a non-empty F = {F : F ⊆ X such that there exist S1, . . . , Sn ∈ S, with is said to be the ﬁlter generated by S. n i=1 Si ⊆ F.} Let F1 and F2 be ﬁlters on a set X. Then F1 is said A6.1.6 Deﬁnitions. to be ﬁner than F2, and F2 is said to be coarser than F1, if F2 ⊆ F1. Further, if F1 = F2 also, then F1 is said to be strictly ﬁner than F2, and F2 is said to be strictly coarser than F1. 623 A6.1.7 Remarks. (i) Let S be a non-empty set of subsets of a non-empty set X, where S has the ﬁnite intersection property. Then the ﬁlter generated by S is the coarsest ﬁlter on X containing the set S. (ii) Let I be an index set and Fi, i ∈ I, a non-empty set of ﬁlters on a set X. Put S = {F : F ∈ Fi, i ∈ I}. If S has the ﬁnite intersection property, then by Proposition A6.1.4, F = {F : F ⊆ X such that there exist S1, . . . , Sn ∈ S, n ∈ N, with n i=1 Si ⊆ F.} is the ﬁlter generated by S, and is the coarsest ﬁlter containing each of the ﬁlters Fi, i ∈ I. So F is the coarsest ﬁlter that is ﬁner than each Fi, i ∈ I. A6.1.8 Deﬁnition. U is said to be an ultraﬁlter on X if no ﬁlter on X is strictly ﬁner than U . Let X be a non-empty set and U a ﬁlter on X. Then Let X be a non-empty set and x0 ∈ X. Let U be the set of A6.1.9 Remark. all subsets of X which contain the given element x0. Then U is an ultraﬁlter. The proof of this is left as an exercise. 624 APPENDIX 6: FILTERS AND NETS A6.1.10 Proposition. then there exists an ultraﬁlter U on X such that U is ﬁner than F0. (Ultraﬁlter Lemma) If F0 is any ﬁlter on a set X, Let F0 denote the set of all ﬁlters on the set X which are ﬁner than the Proof. given ﬂiter F0. Noting Deﬁnition 10.2.1, we see that we can make F0 into a partially ordered set (F0, ) by putting Fi Fj, for Fi, Fj ∈ F0, if Fi ⊆ Fj. Observe that as F0 ∈ F0, F0 is a non-empty set. We shall now apply Zorn’s Lemma 10.2.16. Let F1 be any subset of F0. Then the partial order on F0 induces a partial order on F1. If (F1, ) is a linearly ordered set of ﬁlters, then we claim that there is a ﬁlter F2 which is an upper bound of F1. Deﬁne F2 = {F : F ∈ Fi, Fi ∈ F1}; in other words, F2 is the union of all the ﬁlters in F1. We will show that F2 is a ﬁlter by verifying it satisﬁes Deﬁnition A6.1.1. As the empty set, Ø, is not a member of any of the ﬁlters Fi in F1, Ø /∈ F2. If F1, F2 ∈ F2, F1 ∈ Fi ∈ F1 and F2 ∈ Fj ∈ F1. As F1 is a linearly ordered set, without loss of generality we can assume Fi Fj; that is, Fi ⊆ Fj. So Fi, F2 ∈ Fj. As Fj is a ﬁlter, F1 ∩ F2 ∈ Fj ⊆ F2. Finally, let F ∈ F2 and let G be a subset of X containing F . Then F ∈ Fi ∈ F1. As Fi is a ﬁlter, G ∈ Fi ⊆ F2. So F2 is indeed a ﬁlter. Clearly F2 is an upper bound of F1. Then by Zorn’s Lemma 10.2.16, F0 has a maximal element, U . Suppose F3 is a ﬁlter on X which is strictly ﬁner than U . As U ∈ F0, U is ﬁner than F0. Then F3 is also ﬁner than F0. So F3 ∈ F0. But this contradicts the maximality of U . So our supposition is false, and no ﬁlter is strictly ﬁner than U . So U is indeed an ultraﬁlter and is ﬁner than F0. A6.1.11 Remark. As any ﬁlter ﬁner than a free ﬁlter is a free ﬁlter, it follows from the Ultraﬁlter Lemma A6.1.10 and Example A6.1.3(iii) that free ultraﬁlters exist. The existence of free ultraﬁlters is by no means obvious. As an immediate consequence of the Ultraﬁlter Lemma A6.1.10 and Proposition A6.1.4 we have the next corollary which is, in fact, a generalisation of the Ultraﬁlter Lemma A6.1.10. If S is a non-empty set of subsets of a non-empty set A6.1.12 Corollary. X and S has the ﬁnite intersection property, then there is an ultraﬁlter U on X such that S ⊆ U . 625 If A and B are A6.1.13 Corollary. subsets of X such that A ∪ B ∈ U , then either A ∈ U or B ∈ U . In particular, if A is any subset of X, then either A ∈ U or X \ A ∈ U . Let U be an ultraﬁlter on a set X. Suppose there exist F1, F2 ∈ U such that F1 ∩ A = Ø and F2 ∩ B = Ø, Proof. where A ∪ B ∈ U . Then (F1 ∩ F2) ∩ (A ∪ B) = Ø. But this is a contradiction as F1 ∩ F2 and A ∪ B are members of the ﬁlter U . So, without loss of generality, asssume F ∩ A = Ø, for all F ∈ U . Let S = U ∪ {A}. Then S has the ﬁnite intersection property. By Corollary A6.1.12, there exists an ultraﬁlter U1 such that U ⊆ S ⊆ U1. But this is a contradiction unless U = U1, as the ultraﬁlter U is maximal. Thus A ∈ U , which completes the proof. Corollary A6.1.13 suggests a characterization of ultraﬁlters amongst all ﬁlters. Let F be a ﬁlter on a set X. Then F is an ultraﬁlter A6.1.14 Proposition. on X if and only if it has the property that for every subset A of X, either A ∈ F or (X \ A) ∈ F . Proof. A of X, A ∈ F or X \ A ∈ F . If F is an ultraﬁlter, then Corollary A6.1.13 shows that for every subset Now assume that for every subset A of X, A ∈ F or (X \ A) ∈ F . Suppose F is not an ultraﬁlter. Then there exists a ﬁlter F1 such that F ⊂ F1. So there exists a subset S of X such that S ∈ F1 but S ∈ F . By our assumption, (X \ S) ∈ F . This implies (X \ S) ∈ F1. So both S and X \ S are members of the ﬁlter F1 and S ∩ (X \ S) = Ø. This is a contradiction to F1 being a ﬁlter. So our supposition is false, and F is indeed an ultraﬁlter. 626 APPENDIX 6: FILTERS AND NETS A6.1.15 Corollary. Ui = Ø or there exists an x0 ∈ X such that {x0} ∈ U and Ui∈U Let U be an ultraﬁlter on a set X. Then either Ui = {x0}. Ui∈U Proof. (X \ {x}) ∈ U . If {x} ∈ U , then Ui∈ U Let x be any point in X. By Proposition A6.1.14, either {x} ∈ U or Ui = {x} or Ø, as Ui ⊆ {x}. So either Ui∈ U required. On the other hand, if {x} /∈ U , for all x ∈ X, then (X \ {x}) ∈ U , for all x ∈ X. In this case, Ui ⊆ (X \ {x}) = Ø. Ui∈ U x∈X Let f be a mapping of a set X into a set Y and let A6.1.16 Corollary. U be an ultraﬁlter on X. Then f (U) is an ultraﬁlter on Y if and only if f is surjective. ByExamples A6.1.3(v), f (U) is a ﬁlter on Y if and only if f is surjective. Proof. So let us assume that f is surjective. We shall prove that f (U) is an ultraﬁlter using Proposition A6.1.14. So, let A be any subset of Y . As f is surjective, (X \ f −1(A)) = f −1(Y \ A). Since U is an ultraﬁlter, f −1(A) ∈ U or (X \ f −1(A)) ∈ U . Thus A ∈ f (U) or f (X \ f −1(A)) = f (f −1(Y \ A)) = (Y \ A) ∈ f (U). So f (U) is an ultraﬁlter by Proposition A6.1.14. A6.1.17 Proposition. the set of all ultraﬁlters on X which are ﬁner than F . Then F = Let F be a ﬁlter on a set X and let Ui, i ∈ I, be i∈I Ui. 627 Proof. For each i ∈ I, Ui ⊇ F . So i∈I Ui ⊇ F . Suppose that there exists a subset S of X, such that S /∈ F , but S ∈ Ui, for each i ∈ I. As each Ui is a ﬁlter, (X \ S) /∈ Ui, for each i ∈ I. So (X \ S) /∈ F . Let F = {Fj : j ∈ J} for some index set J . Either (i) (X \ S) ∩ Fj = Ø, for each j ∈ J or (ii) (X \ S) ∩ Fj = Ø, for some j ∈ J . In case (i), {Fj : j ∈ J} ∪ {X \ S} has the ﬁnite intersection property. So by Corollary A6.1.12, there exists an ultraﬁlter U such that U ⊇ ({Fj : j ∈ J}∪{X \S}); that is, (X \ S) ∈ U and F ⊆ U . So U = Ui and (X \ S) ∈ Ui, for some i ∈ I. This is a contradiction, and so case (i) cannot occur. In case (ii), for some j ∈ J , (X \ S) ∩ Fj = Ø and thus Fj ⊆ S. As Fj ∈ F and F is a ﬁlter, this implies S ∈ F . This too is a contradiction, and so case (ii) cannot occur. Thus our supposition is false; that is, S cannot exist. This says that F = i∈I Ui, as required. A6.1.18 Remark. As an immediate consequence of Proposition A6.1.17 we obtain the following: If F is a ﬁlter on a set X and there exists an ultraﬁlter U on X which is strictly ﬁner than F , then there must exist at least one other ultraﬁlter (strictly) ﬁner than F . While ﬁlters on sets are interesting, our focus is on topological spaces. So we now examine the interplay between topologies and ﬁlters via the notion of convergence. 628 APPENDIX 6: FILTERS AND NETS Let (X, τ ) be a topological space, x ∈ X, and F a A6.1.19 Deﬁnition. ﬁlter on the set X. Then the ﬁlter F is said to converge on (X, τ ) to x, and x is said to be a limit in (X, τ ) of F , denoted by x ∈ lim F or F → x, if each neighbourhood in (X, τ ) of x is a member of F . The set of all limit points in (X, τ ) of F is denoted by lim F . A6.1.20 Remark. neighbourhood ﬁlter, Nx of x converges on (X, τ ) to x. Let (X, τ ) be a topological space and x ∈ X. Then the Let (X, τ ) be a topological space and x ∈ X. A ﬁlter F A6.1.21 Remark. on the set X converges to x if and only if F is ﬁner than the neighbourhood ﬁlter, Nx, of x. [Verify this.] A6.1.22 Proposition. Let (X, τ ) be a topological space. Then the following are equivalent: (i) (X, τ ) is a Hausdorﬀ space; (ii) Every ﬁlter F on (X, τ ) converges to at most one point; (iii) Every ultraﬁlter U on (X, τ ) converges to at most one point. Proof. Exercise. Let (X, τ ) be a topological space. Then (X, τ ) is A6.1.23 Proposition. compact if and only if for every ﬁlter F on (X, τ ) there is a ﬁlter F1 which is ﬁner than F and converges. 629 Proof. Recall that, by Proposition 10.3.2, a topological space is compact if and only if for every family S of closed subsets with the ﬁnite intersection property, Si∈S Si =
Ø. Assume that (X, τ ) is compact and let F be any ﬁlter on (X, τ ). Then F has the ﬁnite intersection property. Put G = {F : F ∈ F}. Then G has the ﬁnite intersection property too. As (X, τ ) is compact, there exists a point x0 ∈ X, such Fi∈F Fi. So if Nx0 ∈ Nx0, the neighbourhood ﬁlter in (X, τ ) of x0, that x0 ∈ then Nx0 ∩ Fi = Ø, for all Fi ∈ F . Thus F and Nx0 are ﬁlters on (X, τ ) such that S = {S : S ∈ Nx0 or S ∈ F} has the ﬁnite intersection property.. So by Remark A6.1.7(ii), there exists a ﬁlter F1 on (X, τ ) such that F1 is ﬁner than both F and Nx0. As F1 is ﬁner than the neighbourhood ﬁlter Nx0, Remark A6.1.21 says that F1 → x0. As the ﬁlter F1 is also ﬁner than F , this proves the required result. Conversely, assume that for every ﬁlter on (X, τ ) there is a ﬁlter which is ﬁner than it which converges. Let S be a family of closed subsets of (X, τ ) with the ﬁnite intersection property. By Proposition A6.1.4 (ii), there is a ﬁlter F on (X, τ ) which contains S. So by assumption, there exists a point x0 in X and a ﬁlter F1 ⊇ F such that F1 → x0. So by Remark A6.1.21, F1 ⊇ Nx0, the neighbourhood ﬁlter of x0. Thus for each Nx0 ∈ Nx0, Nx0 ∩ F = Ø, for each F ∈ F1. In particular, Nx0 ∩ Si = Ø, for every Si ∈ S. Therefore x0 ∈ Si∈ S Si. So (X, τ ) is compact. 630 APPENDIX 6: FILTERS AND NETS While Proposition A6.1.23 is a nice characterization of compactness using ﬁlters, its Corollary A6.1.24 is a surprisingly nice characterization of compactness using ultraﬁlters. A6.1.24 Corollary. compact if and only if every ultraﬁlter on (X, τ ) converges. Let (X, τ ) be a topological space. Then (X, τ ) is Let U be any ultraﬁlter on (X, τ ). By Proposition A6.1.23, there is a Proof. ﬁlter F1 ⊇ U which converges. But as U is an ultraﬁlter, U = F1. So U converges. Conversely, assume that every ultraﬁlter on (X, τ ) converges. Let F be any ﬁlter on (X, τ ). Then, by the Ultraﬁlter Lemma A6.1.10, there is an ultraﬁlter U ⊇ F . By assumption U converges; that is, there is a ﬁlter ﬁner than F which converges, which completes the proof. We now present a proof of Tychonoﬀ ’s Theorem 10.3.4 using ultraﬁlters. 631 A6.1.25 Theorem. be any family of topological spaces. Then if each (Xi, τ i) is compact. (Tychonoﬀ ’s Theorem) Let {(Xi, τ i) : i ∈ I} i∈I (Xi, τ i) is compact if and only Proof. Xi = pj( i∈I (Xi, τ i) is compact, then each (Xi, τ i), i ∈ I, is compact as If i∈I Xi) and a continuous image of a compact space is compact. Conversely, assume each (Xi, τ i), i ∈ I is compact. Let U be any ultraﬁlter on i∈I (Xi, τ i). By Corollary A6.1.24, it suﬃces to prove that U converges. By Lemma A6.1.16, pi(U) is an ultraﬁlter on (Xi, τ i), for each i ∈ I. As (Xi, τ i) is compact, Corollary A6.1.24 implies that pi(U) converges in (Xi, τ i), for each i ∈ I. Noting Proposition A6.1.22, we see that pi(U) may converge to more than one point if (Xi, τ i) is not Hausdorﬀ. Using the Axiom of Choice, for each i ∈ I, choose xi ∈ Xi such that pi(U) → xi. Let Oj be any open neighbourhood in (Xj, τ j) of xj. As pj(U) → xj, Oj ∈ Xi] /∈ pj(U). Since pj(U) is a ﬁlter, (Xj \Oj) /∈ pj(U). Therefore [(Xj \Oj)× U . By Proposition A6.1.14, this implies that [Oj × Xi] ∈ U . By Deﬁnition i∈I\{j} i∈I\{j} A6.1.1(ii) this implies that if j1, j2, . . . , jn ∈ I, and Ojk in (Xjk, τ jk) of xjk , k = 1, 2, . . . , n, then is any open neighbourhood [Oj1 × Oj2 × . . . Ojn × Xi] ∈ U. i∈I\{j1,...,jn} As every neighbourhood, Nx, in i∈I set of the form [Oj1 × Oj2 × . . . Ojn × Xi of x = i∈I xi must contain a basic open i∈I\{j1,...,jn} Xi], it follows that Nx ∈ U , because U is a ﬁlter. So U ⊇ Nx, and hence U → x, as required. 632 APPENDIX 6: FILTERS AND NETS There are three 15 minute YouTube videos which, together with Appendix 1, would help put Remark A6.1.26 into context. The videos are:- Topology Without Tears – Inﬁnite Set Theory – Video 2a – http://youtu.be/9h83ZJeiecg, Video 2b – http://youtu.be/QPSRB4Fhzko, & Video 2c – http://youtu.be/YvqUnjjQ3TQ A6.1.26 Remark. It is appropriate for us to say a few words about Set Theory, on which our study of Topology rests. The standard axioms for Set Theory are known as the Zermelo-Fraenkel (ZF) axioms. These are named after the mathematicians, Ernst Zermelo [1871–1953] and Abraham Halevi Fraenkel37[1891–1965]. In 1904, Zermelo formulated the Axiom of Choice (AC), which says: For any set {Xi : i ∈ I} of non-empty sets, we can choose a member from each Xi. More formally this says: For any set {Xi : i ∈ I} of non-empty sets, there exists a function f : I → Xi such that f (x) ∈ Xi for every i ∈ I. i∈I In language more familiar to readers of this book, we can state the Axiom of Choice as follows: For any set {Xi : i ∈ I} of non-empty sets, Zermelo said that the Axiom of Choice is an unobjectionable logical principle. It i∈I Xi is non-empty. is known that the Axiom of Choice is not implied by the Zermelo-Fraenkel axioms, but is consistent with them; that is, AC is independent of, but consistent with, ZF. During much of the twentieth century, AC was controversial. Mathematicians were divided on whether AC is true or false or whether it should be assumed [that is, added to the ZF axioms] and used. Jan Mycielski [300] relates an anecdote which 37Fraenkel was the ﬁrst Dean of Mathematics at the Hebrew University of Jerusalem. 633 demonstrates this very well: The mathematician Alfred Tarski (1902–1983) tried to publish his theorem proving the equivalence between AC and “every inﬁnite set A has the same cardinality as the product set A × A” in the oldest extant mathematics journal in the world, Comptes Rendus, but the mathematicians Maurice René Fréchet (1878–1973) and Henri-Léon Lebesgue (1875–1941) refused to present it. Fréchet wrote that an implication between two well known true propositions is not a new result, and Lebesgue wrote that an implication between two false propositions is of no interest. In this book we do not hesitate to use the Axiom of Choice. The Axiom of Choice is equivalent to Zorn’s Lemma 10.2.16 and to the Well-Ordering Theorem [WOT] 10,2,15 (which says that every set can be well-ordered). It is also equivalent to Tychonoﬀ ’s Theorem which says that any product of compact spaces is compact. However, if you examine the above proof of Tychonoﬀ ’s Theorem, you will see that the special case of Tychonoﬀ ’s Theorem which says that any product of compact Hausdorﬀ spaces is compact Hausdorﬀ, does not require the Axiom of Choice but only the Ultraﬁlter Lemma, as pi(U) converges to a unique point in a Hausdorﬀ space and so no choice is needed. But we did use Zorn’s Lemma in our proof of the Ultraﬁlter Lemma. While the Ultraﬁlter Lemma is implied by the Axiom of Choice, it is not equivalent to the Axiom of Choice, it is in fact weaker. In summary, the following four statements are equivalent to each other: (i) Axiom of Choice (ii) Zorn’s Lemma (iii) Well-Ordering Theorem and (iv) Tychonoﬀ ’s Theorem. Each of these implies, but is not implied by, the Ultraﬁlter Lemma. Further, none of these is implied by the Zermelo-Fraenkel axioms. For deeper discussion of the Axiom of Choice, see Rubin and Rubin [348], Rubin and Rubin [349] and Herrlich [176]. [The book Howard and Rubin [199] is a survey of research done during the last 100 years on the Axiom of Choice and its consequences, and is updated on Howard [198].] 634 APPENDIX 6: FILTERS AND NETS Let f be a surjective mapping of a topological space A6.1.27 Proposition. (X, τ ) onto a topological space (Y, τ 1), and for each x ∈ X, let Nx and Nf (x) denote, respectively, the ﬁlter of neighbourhoods of x ∈ (X, τ ) and the ﬁlter of neighbourhoods of f (x) ∈ (Y, τ 1). Then the following are equivalent: (i) f is continuous; (ii) for every x ∈ X, Nf (x) ∈ Nf (x) =⇒ f −1(Nf (x)) ∈ Nx; (iii) for every x ∈ X, f (Nx) ⊇ Nf (x); (iv) for every x ∈ X and every ﬁlter F on (X, τ ), F → x =⇒ f (F) → f (x); that is, F ⊇ Nx =⇒ f (F) ⊇ Nf (x); (v) for every x ∈ X and every ultraﬁlter U on (X, τ ), U → x =⇒ f (U) → f (x); that is, U ⊇ Nx =⇒ f (U) ⊇ Nf (x). Proof. We begin by noting that Remark A6.1.21 says that a ﬁlter F converges to a point x if and only if F is ﬁner than Nx. Clearly (i) ⇔ (ii) ⇔ (iii) and (iii) =⇒ (iv) =⇒ (v). We shall complete the proof of the Proposition by showing that (v) =⇒ (ii). Assume (v) is true. Let x ∈ X and U be any ultraﬁlter ﬁner than Nx. Let Nf (x) ∈ Nf (x). Then Nf (x) ∈ f (U). By Examples A6.1.13(v), f −1(Nf (x)) ∈ U . By Proposition A6.1.17, Nx equals the intersection of all ultraﬁlters containing it. As f −1(Nf (x)) is in each such ultraﬁlter, f −1(Nf (x)) ∈ Nx. Thus (v) =⇒ (ii). The same kind of argument used above yields the next Proposition. 635 Let (X, τ ) be a topological space and O a subset If Nx denotes the neighbourhood ﬁlter of any point x ∈ X, then the A6.1.28 Proposition. of X. following are equivalent: (i) O is an open set in (X, τ ); (ii) for every x ∈ O, there exists an Nx ∈ Nx such that Nx ⊆ O; (iii) for every x ∈ O, O ∈ Nx; (iv) for every x ∈ O, and every ﬁlter Fx → x, O ∈ Fx; (v) for every x ∈ O, and every ultraﬁlter Ux → x, O ∈ Ux. Proof. Exercise. The next corollary follows immediately. Let (X, τ ) be a topological space and C a subset of X. A6.1.29 Corollary. If Nx denotes the neighbourhood ﬁlter of any point x ∈ X, then the following are equivalent: (i) C is a closed set in (X, τ ); (ii) for every x ∈ (X \ C), there exists an Nx ∈ Nx such that Nx ⊆ (X \ C); (iii) for every x ∈ (X \ C), (X \ C) ∈ Nx; (iv) for every x ∈ (X \ C), and every ﬁlter Fx → x, (X \ C) ∈ Fx; (v) for every x ∈ (X \ C), and every ultraﬁlter Ux → x, (X \ C) ∈ Ux. 636 APPENDIX 6: FILTERS AND NETS Let X be a non-empty set and for each x ∈ X, let A6.1.30 Proposition. Sx be a non-empty set of ﬁlters on X such that x ∈ Fxi, for each Fxi ∈ Sx. Let τ be the set of subsets of X deﬁned as follows: O ∈ τ if for each x ∈ O, O ∈ Fxi, for every ﬁlter Fxi ∈ Sx. Then τ is a topology on X. Further, for each x ∈ X and each ﬁlter Fxi ∈ Sx, Fxi → x in (X, τ ) and Nx = Fxi is the nei
ghbourhood ﬁlter at x in (X, τ ). Fxi∈Sx Proof. Exercise. Let X be a non-empty set and τ 1 a topology on the A6.1.31 Corollary. set X. Further let Sx be the set of all ultraﬁlters which converge to x on (X, τ 1). Let τ be the topology deﬁned from Sx, x ∈ X, as in Proposition A6.1.30. Then τ = τ 1. Proof. Exercise. A6.1.32 Remark. We have now seen that a topology τ on a set X determines convergent ﬁlters and convergent ultraﬁlters, and conversely a set of ﬁlters or ultraﬁlters determines a topology. We have also seen how ﬁlters or ultraﬁlters can be used to deﬁne continuous functions and compactness. And we have seen that ultraﬁlters can be used to give an elegant and short proof of Tychonoﬀ ’s Theorem. Exercises A6.1 637 1. Find all ﬁlters on each of the following sets: (i) {1, 2}; (ii) {1, 2.3}; and (iii) {1, 2, 3, 4}. 2. Let X be a set and for some index set I, let {Fi : i ∈ I} be ﬁlters on the set X. Prove the following statements: (i) F = i∈I Fi is a ﬁlter on the set X; that is, the intersection of any (ﬁnite or inﬁnite) number of ﬁlters on a set X is a ﬁlter; (ii) F = i∈I Fi = i∈I Fi : Fi ∈ Fi. [Hint: This curious looking fact follows from the fact that each Fi is a ﬁlter.] 3. Verify that a ﬁlter F on a ﬁnite set is not a free ﬁlter. 4. Prove the statements in Examples A6.1.3 (iii) and (v). 5. Let X be a set with at least two points, τ a topology on X other than the indiscrete topology, and F = τ \ {Ø}. If F is a ﬁlter, then (X, τ ) is (i) connected; (ii) extremally disconnected (that is, the closure of every open set is open); (iii) not metrizable; (iv) not Hausdorﬀ; and (v) not a regular space. 6. Let F be a ﬁlter on a set X such that for some F ∈ F , F = X. If x0 ∈ X but x0 ∈ F , show that the set X \ {x0} ∈ F . Using this, prove that every free ﬁlter on a set X is ﬁner than the Fréchet ﬁlter on X. 7. Let S = {N \ {n} : n ∈ N}. Find the ﬁlter on N which is generated by S. 8. (i) Let X be a set and G a family of subsets of X. Prove that there exists a ﬁlter F such that F ⊇ G if and only if G has the ﬁnite intersection property. 638 APPENDIX 6: FILTERS AND NETS (ii) If G is a set of subsets of X and G has the ﬁnite intersection property, prove that the ﬁlter generated by G is the coarsest ﬁlter F which contains G. 9. Let S be a subset of a set X and F be a ﬁlter on X. Find a necessary and suﬃcient condition for F1 = {F ∩ S : F ∈ F} to be a ﬁlter on the set S. 10. Find all ultraﬁlters on each of the following sets: (i) {1, 2}; (ii) {1, 2.3}; and (iii) {1, 2, 3, 4}. 11. Prove the statement in Remark A6.1.9. 12. Let F be an ultraﬁlter on a set X. If A and B are subsets of X we have seen that A ∪ B ∈ F implies that A ∈ F or B ∈ F . Extend this result in a natural way to sets A1, A2, . . . , An where A1 ∪ A2 ∪ . . . An ∈ F . 13. Prove the statement in Remark A6.1.21. 14. Prove the statements in Proposition A6.1.22. 15. Prove the statements in Proposition A6.1.28. 16. Let (X, τ ) be a topological space, x ∈ X and S a subset of X. Prove that x is a limit point of S if and only if S \ {x} is a member of some ﬁlter F which converges to x. 639 17. Let (X, τ ) be a topological space. Prove the following statements: (i) (X, τ ) is a T0-space if and only if for each x, y ∈ X, such that x = y, there exists a ﬁlter Fxy on (X, τ ) such that either x ∈ lim Fxy and y /∈ lim Fxy or y ∈ lim Fxy and x /∈ lim Fxy; (ii) (X, τ ) is a T1-space if and only if for each x, y ∈ X, such that x = y, there exists ﬁlters Fxy and Fyx on (X, τ ) such that (a) x ∈ lim Fxy and y /∈ lim Fxy and (b) y ∈ lim Fyx and x /∈ lim Fyx; (iii)* (X, τ ) is a Hausdorﬀ space if and only if each ﬁlter F on (X, τ ) has at most one limit point. [Hint. If (X, τ ) is not a Hausdorﬀ space, let x, y ∈ X be such that x = y but each open neighbourhood of x intersects each open neighbourhood of y. Let Fxy consist of all the subsets of X which contain sets of the form Nx ∩ Ny, for some neighbourhood Nx of x and neighbourhood Ny of y. Show that Fxy is a ﬁlter on (X, τ ), x ∈ lim Fxy and y ∈ lim Fxy. So Fxy has more than one limit point. Conversely, let (X, τ ) be a Hausdorﬀ space and x, y ∈ X with x = y. Then there exists neighbourhoods Nx and Ny of x and y, respectively, such that Nx ∩ Ny = Ø. Note that Nx and Ny cannot both be members of any ﬁlter.] 18. Let τ 1 and τ 2 be topologies on a set X. Prove that the topology τ 1 is ﬁner than the topology τ 2 if and only if every convergent ﬁlter on (X, τ 1) also converges to the same point(s) on (X, τ 2). 19. For each i in an index set I, let (Xi, τ i) be a topological space and let (Xi, τ i), their product space with the product topology. Let pi (X, τ ) = i∈I be the projection mapping of (X, τ ) onto (Xi, τ i), for each i ∈ I. Further, let F be a ﬁlter on (X, τ ). Prove that the ﬁlter F converges to a point x ∈ X if and only if the ﬁlter pi(F) converges to pi(x) in (Xi, τ i), for each i ∈ I. 20. Prove the statements in Proposition A6.1.30. 21. Prove the statement in Proposition A6.1.31. 640 APPENDIX 6: FILTERS AND NETS A6.2 Filterbases We have seen that it is often more convenient to describe a basis for a topology than the topology itself. For example in R, Rn for n > 1, metric space topologies, and product topologies there are elegant descriptions of their bases but less than elegant descriptions of the topologies themselves. Similarly, it is often more covenient to describe a ﬁlter basis than the ﬁlter itself. Further, we will see that it it is easy to relate convergent ﬁlter bases to convergent sequences. Let X be a non-empty set and G a set of non-empty A6.2.1 Deﬁnition. subsets of X. Then G is said to be a ﬁlterbase (or a ﬁlter basis or a ﬁlterbasis or a ﬁlter base) if G1, G2 ∈ G implies that there exists a G3 ∈ G such that G3 ⊆ G1 ∩ G2. A6.2.2 Remarks. Let X be a non-empty set and G a ﬁlterbase on X. Then (i) clearly Ø /∈ G; (ii) every ﬁlter on X is also a ﬁlterbase on X; (iii) G has the ﬁnite intersection property; (iv) the set F = {F ⊆ X : F ⊇ G, G ∈ G} is a ﬁlter, indeed it is the coarsest ﬁlter, containing G and it is called the ﬁlter generated by G; (v) if X is a set and x ∈ X, then G = {x} is a ﬁlterbase on X; (vi) if X has more than one point, then F = {{x} : x ∈ X} is not a ﬁlterbase or a ﬁlter. Indeed, if x and y are distinct points of X, then a ﬁlter or ﬁlterbasis F cannot contain both of the sets {x} and {y}. (vii) If A is a set of non-empty subsets of X and A has the ﬁnite intersection property, then there exists a ﬁlterbase containing A. [Verify this.] 641 A6.2.3 Examples. (i) Let (X, τ ) be a topological space and x any point in X. Let Nx be the set of all neighbourhoods of x in (X, τ ). Then Nx is clearly a ﬂiterbase. It is the neighbourhood ﬁlterbase of x. (ii) Let [0,1] be the closed unit interval with the Euclidean topology. Let N = {[0, 1/n) : n ∈ N}. Then N is a ﬁlterbase of open neighbourhoods of the point 0. (iii) Let X be a set and xi, i ∈ N, a sequence of points in X. For each n ∈ N, let Gn = {xj : j ∈ N and j n}. So each Gn is a subset of X and we deﬁne G = {Gn : n ∈ N}. Then G is a ﬁlterbase on the set X. We call G the ﬁlterbase determined by the sequence xi, i ∈ N,. Let G1 and G2 be ﬁlterbases on a set X. Then (i) G1 A6.2.4 Deﬁnitions. is said to be ﬁner than G2, (ii) G2 is said to be coarser than G1, and (iii) G1 is said to be a reﬁnement of G2 if for each set G2 ∈ G2, there exists a G1 ∈ G1 such that G1 ⊆ G2. If G1 is both coarser and ﬁner than G2, then G1 and G2 are said to be equivalent ﬁlterbases. A6.2.5 Example. topology. Let G1 = {[0, 1 G2 are easily seen to be equivalent ﬁlterbases. Let [0,1] be the closed unit interval with the Euclidean 2n+1) : n ∈ N}. Then G1 and 2n) : n ∈ N} and G2 = {[0, 1 Consider the sequence 1, 1 6.2.6 Example. n, . . . , and let G1 be the ﬁlterbase determined by this sequence as in Example 6.2.3 (iii); that is, n, 1 G1 = {G1n : n ∈ N}, where G1n = { 1 n+1, 1 n+2, . . . }. Let G2 be the ﬁlterbase determined by the sequence 1, 1 9, . . . , 1 3n, . . . . So G2 = {G2n : n ∈ N}, where G2n = { 1 1 3n, 3(n+2), . . . }. Then each G2n ⊂ G1n and so G2 is a reﬁnement 3(n+1), of G1. 3, . . . , 1 3, 1 6, 1 2, 1 1 642 APPENDIX 6: FILTERS AND NETS Let (X, τ ) be a topological space and G a ﬁlterbase A6.2.7 Deﬁnition. on X. If x ∈ X, then G is said to converge to x, denoted by G → x and x is said to be a limit point of the ﬁlterbase G, denoted by x ∈ lim G, if for every neighbourhood U of x there is a G ∈ G such that G ⊆ U . 3, . . . , 1 A6.2.8 Example. On the Euclidean space R consider the sequence 1, 1 where n ∈ N. Let G be the ﬁlterbase determined by this sequence, as in Example A6.2.3(iii). Then clearly G → 0. 2, 1 n, . . . , Let (X, τ ) be any topological space and x any point in A6.2.9 Example. X. Let Nx be the set of all neighbourhoods of x in (X, τ ); that is, Nx is the neighbourhood ﬂiterbase at x of Example A6.2.3(i). Clearly Nx → x A6.2.10 Remark. ﬁlterbases on X. Further let a be a point in X and G1 a reﬁnement of G2. G2 → a, then clearly G1 → a. Let (X, τ ) be a topological space and let G1 and G2 be If Let (X, τ ) be a topological space, S a subset of X, A6.2.11 Proposition. and a a point in X \ S. Then a is a limit point of the set S if and only if there exists a ﬁlterbase F such that F → a in (X, τ ) where each F ∈ F satisﬁes F ⊆ S. If a is a limit point of S, then each neighbourhood N of a must Proof. intersect the set S nontrivially. Let FN = N ∩ S. Let N denote the ﬁlterbase of neighbourhoods in (X, τ ) of a. We claim that F = {FN : N ∈ N } is a ﬁlterbase. Each FN is non-empty. Further if N1, N2 ∈ N , then the set N3 = N1 ∩ N2 ∈ N . But then as a is a limit point of S and N3 is a neighbourhood of a, we also have , and so FN is FN3 = N3 ∩ S = Ø and FN3 = N3 ∩ S = (N1 ∩ N2) ∩ S = FN1 ∩ FN2 indeed a ﬁlterbase. Further, from the very deﬁnition of F , we see that F → a, as required. Conversely, let F1 be a ﬂiterbase such that F1 → a ∈ X, and each F ∈ F1, satisﬁes F ⊆ S. Let N be any neighbourhood of a. As F1 → a, this implies that N ⊇ F1, for some F1 ∈ F
1. Since F1 ∈ F1, F1 ⊆ S, and thus N ∩ S = Ø. Hence a is a limit point of S, as required. 643 Let (X, τ ) be a topological space and S a subset of A6.2.12 Corollary. X. The following two properties are equivalent: (i) S is a closed set in (X, τ ); (ii) Let F be a ﬁlterbase on (X, τ ) such that F ∈ F implies F ⊆ S. Further, let a ∈ X be such that F → a. Then a ∈ S for every such F and a. A6.2.13 Remark. We know that if (X, τ ) is a metrizable topological space, then a subset S of X is closed if and only if each point a in X such that a is a limit point of a sequence s1, s2, . . . , sn, . . . where each sn ∈ S, satisﬁes a ∈ S. Noting Example A6.2.3 which says that each sequence determines a ﬁlterbase, we see that Corollary A6.2.12 is a generalization of this metrizable space result to arbitrary topological spaces; that is, ﬁlterbases for arbitrary topological spaces fulﬁl the same role that sequences do for metrizable spaces. Let G be a ﬁlterbase on the topological space (X, τ ) A6.2.14 Proposition. and a ∈ X. Further, let F be the ﬁlter generated by the ﬁlterbase G. Then a ∈ lim G if and only if a ∈ lim F . Proof. Exercise. Let (X, τ ) be a topological space and G a ﬁlterbase A6.2.15 Deﬁnition. on X. A point x ∈ X is said to be a cluster point of the ﬁlterbase, G, if for each G ∈ G and each neighbourhood U of x, G ∩ U = Ø. 644 APPENDIX 6: FILTERS AND NETS A6.2.16 Remarks. (i) Every limit point of a ﬁlterbase is a cluster point of that ﬁlterbase. (ii) Consider the sequence 0, 1, 0, 2, 0, 3, . . . , 0, n, . . . , where n ∈ N, in the Euclidean space R. Let G be the ﬁlterbase determined by this sequence as in Example A6.2.3 (iii). Then 0 is clearly a cluster point of G, but 0 is not a limit point of G. (Exercise: Verify this.) Let (X, τ ) be a topological space, a a point in X, A6.2.17 Proposition. and G a ﬁlterbase on (X, τ ). If G has a as a cluster point, then there is a ﬁlterbase G1 on (X, τ ) such that G1 is a reﬁnement of G and a is a limit point of G1. Proof. the set G1 = {GU : U a neighbourhood of a, G ∈ G}. For each neighbourhood U of a and each G ∈ G, let GU = G ∩ U . Deﬁne G1 = G (X, τ ), and G Firstly we have to verify that G1 is a ﬁlterbase. Let G1 and G2 be in G1. Then 2 ∩ U2, where U1 and U2 are neighbourhoods of a in 1 ∩ U1 and G2 = G 1 and G 2 are in G. Then 1 ∩ U1) ∩ (G G1 ∩ G2 = (G As G is a ﬁlterbase, there exists a G U2 are neighbourhoods of a, U3 = U1 ∩ U2 is also a neighbourhood of a. As G G indeed a ﬁlterbase. 3 ∩ U3 ∈ G1. Thus G1 ∩ G2 contains the set G 1 ∩ G 2 ∩ U2) = (G 3 ∈ G such that G 2. Since U1 and 3 ∈ G, 3 ∩ U3, which is in G1. So G1 is 2) ∩ (U1 ∩ U2). 1 ∩ G 3 ⊆ G By the deﬁnition of G1, it is a reﬁnement of G. Clearly, also by the deﬁnition of G1, G1 → a; that is a is a limit point of G1. Let (X, τ ) be a topological space, a a point in X, A6.2.18 Proposition. and G a ﬁlterbase on (X, τ ). If G1 is a ﬁlterbase which is a reﬁnement of G and a is a cluster point of G1, then a is a cluster point of G. 645 Proof. neighbourhood of a in (X, τ ). We need to show that G ∩ U = Ø. To prove that a is a cluster point of G, let G ∈ G and let U be any As G1 is a reﬁnement of G, there exists a G1 ∈ G1 such that G1 ⊆ G. Now a is a cluster point of G1 implies that G1 ∩U = Ø. So G∩U ⊇ G1 ∩U = Ø, as required. Let (X, τ ) be a topological space, a a point in X, and A6.2.19 Corollary. G a ﬁlterbase on (X, τ ). Then a is a cluster point of G if and only if there is a ﬁlterbase G1 which is a reﬁnement of G such that a is a limit point of G1. Proof. This is an immediate consequence of Propositions A6.2.17, A6.2.18 and Remark A6.2.16(i). We now show that ﬁlterbases can be used to characterize compact topological spaces. 646 APPENDIX 6: FILTERS AND NETS A6.2.20 Proposition. if every ﬁlterbase on (X, τ ) has a cluster point. A topological space (X, τ ) is compact if and only Firstly assume that (X, τ ) is a compact space and G is a ﬁlterbase Proof. on (X, τ ). Then the set G = {G ∈ G} has the ﬁnite intersection property by Remarks A6.2.2(iii). Therefore the set {G : G ∈ G} of closed sets also has the ﬁnite intersection property. As (X, τ ) is compact, Proposition 10.3.2 implies that there exists a point a ∈ G. So if U is any neighbourhood of a, U ∩ G = Ø. Thus we G∈G see that a is a cluster point of the ﬁlterbase G. To prove the converse, let (X, τ ) be a topological space for which every ﬁlterbase has a cluster point. We shall prove compactness using Proposition 10.3.2. Let S be any set of closed subsets of (X, τ ) such that S has the ﬁnite intersection property. We shall prove that S∈S S = Ø. Deﬁne A to be the set of all ﬁnite interesections of members of S. It suﬃces to prove that A∈A A = Ø. Now by Remark A6.2.2(vii), A is a ﬁlterbase. So by our assumption, there exists a cluster point x of the ﬁlterbase A; that is, if V is any neighbourhood of x, then V ∩ A = Ø, for each A ∈ A. This implies that x is a limit point of the set A. Noting that each A ∈ A, being a ﬁnite intersection of closed sets, is a closet set, this implies that x ∈ A, for each A ∈ A. Hence A∈A A = Ø, as required. A6.2.21 Corollary. every ﬁlterbase on (X, τ ) has a reﬁnement which has a limit point. A topological space (X, τ ) is compact if and only if Next we show that ﬁlterbases can be used to characterize continuous mappings. 647 Let (X, τ ) and (Y, τ 1) be topological spaces and A6.2.22 Proposition. f a continuous mapping of (X, τ ) into (Y, τ 1). If F is a ﬁlterbase on (X, τ ), then f (F) is a ﬁlterbase on (Y, τ 1). Further, if x ∈ X and F → x, then f (F) → f (x). Firstly we show that f (F) is a ﬁlterbase on (Y, τ 1). As F is a ﬁlterbase, Proof. every set in F is non-empty and so each set f (F) is also non-empty. Now let G1, G2 ∈ f (F). Then there exists F1, F2 ∈ F such that f (F1) = G1 and f (F2) = G2. As F is a ﬁlterbase, by Deﬁnition A6.2.1 there exists a non-empty set F3 ∈ F , such that F3 ⊆ F1 ∩ F2. This implies f (F3) ⊂ f (F1) ∩ f (F2); that is, the set G3 = f (F3) ∈ f (F) and G3 ⊆ G1 ∩ G2. So f (F) is indeed a ﬁlterbase on (Y, τ 1. Now we know that f is continuous and F → x. Let U be any neighbourhood of f (x) in (Y, τ 1). As f is continuous, f −1(U ) is a neighbourhood of x in (X, τ ). Since F → x, by Deﬁnition 6.2.7 there exists F ∈ F , such that F ⊆ f −1(U ). So f (F ) ∈ f (F) and f (F ) ⊆ U . Thus f (F) → f (x), as required. We now state and prove the converse of Proposition A6.2.22. Let (X, τ ) and (Y, τ 1) be topological spaces and f A6.2.23 Proposition. a mapping of (X, τ ) into (Y, τ 1). If for each x ∈ X and each ﬁlterbase F → x, the ﬁlterbase f (F) → f (x), then f is a continuous mapping Let U be any open set in (Y, τ 1). We are required to show that the set Proof. f −1(U ) is an open set in (X, τ ). Let x be any point in the set f −1(U ) and Nx the neighbourhood ﬁlter at x in (X, τ ). By Example A6.2.9, Nx → x. By assumption, f (Nx) → f (x). As U is open, it is a neighbourhood of f (x). So there exists a G ∈ f (Nx) such that G ⊆ U . However, G = f (N ), where N ∈ Nx and thus f (N ) ⊆ U . Hence the neighbourhood N of x satisﬁes N ⊆ f −1(U ). So we have that the set f −1(U ) contains a neighbourhood of x, for each x ∈ f −1(U ). Thus f −1(U ) ia an open set, which completes the proof that f is continuous. 648 APPENDIX 6: FILTERS AND NETS Exercises A6.2 1. Let F be a ﬁlterbasis on a set X and G be a ﬁlterbasis on a set Y . Prove that F × G = {F × G : F ∈ F, G ∈ G} is a ﬁlterbasis on the product set X × Y . 2. Prove the statement in Remark A6.2.10. 3. Verify the claim in Remarks A6.2.12(ii) that 0 is a cluster point but not a limit point of G. 4. Prove Proposition A6.2.14. 5. Let (X, τ ) be a topological space and U a ﬁlterbase on (X, τ ). Then U is said to be an ultraﬁlterbase (or an ultraﬁlterbasis or an ultraﬁlter base or ultraﬁlter basis) if the ﬁlter that it generates is an ultraﬁlter. Prove that U is an ultraﬁlterbase if and only if for each set S ⊆ X there exists an F ∈ U such that S ⊇ F or X \ S ⊇ F . 6. (i) Verify Remarks A6.2.16(ii). (ii) In the example in Remarks A6.2.16(ii), ﬁnd a reﬁnement of the ﬁlterbase G which has 0 as a limit point. 7. Verify Corollary A6.2.20. 8. Prove that a metrizable topological space (X, τ ) is compact if and only if every sequence of points in X has a subsequence converging to a point of X. 9. Let (X, τ ) be a topological space. Prove that (X, τ ) is Hausdorﬀ if and only if every ﬁlterbasis, G , on (X, τ ) converges to at most one point. [Hint: See Exercises A6.1 #9(iii).] 10. Let G be a ﬁlterbasis on the Euclidean space R. Verify that G → 0 if and only if for every ε > 0 there exists a set G ∈ G such that x ∈ G implies \|x\| < ε. 11.* Let (Xi, τ i) , for i ∈ I, be a set of topological spaces and G a ﬁlterbasis on the i∈I (Xi, τ i) is a limit i∈I (Xi, τ i). Prove that the point x ∈ product space point of G if and only if φi(G) → φi(x), for each i ∈ I, where φi is the projection mapping of i∈I (Xi, τ i) onto (Xi, τ i). A6.3 Nets38 649 In 1922 E.H. Moore and H.L. Smith in their paper, Moore and Smith [289], introduced the notion of a net as a generalization of a sequence. As we shall see, this notion is equivalent to the more elegant approach of ﬁlters introduced in 1937 by Henri Cartan (Cartan [69], Cartan [70]). A partial order, , on a set was deﬁned in Deﬁnitions 10.2.1. A set with a partial order is called a partially ordered set. §10.2 gives a number of examples of partially ordered sets. A partially ordered set (D, ) is said to be a directed A6.3.1 Deﬁnition. set if for any a ∈ D and b ∈ D, there exists a c ∈ D such that a c and b c. A6.3.2 Examples. order are directed sets. It is easily seen that N, Z, Q, and R with the usual partial Let (X, τ ) be a topological space and a a point in X. Let A6.3.3 Example. D be the set of all neighbourhoods of the point a. Put a partial order on D by D1 D2 if D2 ⊆ D1, where the sets D1, D2 ∈ D. As the intersection of two neighbourhoods of a is a neighbourhood of a, it follows that D1 ∩ D2 ∈ D, and D1 ∩ D2 D1 and D1 ∩ D2 D2, Thus (D, ) is a di
rected set. A6.3.4 Deﬁnition. set. Then a function φ : D → X is said to be a net in the space (X, τ ). Let (X, τ ) be a topological space and (D, ) a directed It is often convenient to write the net φ above as {xα} where φ(α) = xα ∈ X, α ∈ D. 38There are two 14 minute YouTube videos which provide an excellent introduction to this section. They are:- Topology Without Tears – Sequences and Nets – Video 3a – http://youtu.be/wXkNgyVgOJE & Video 3b – http://youtu.be/xNqLF8GsRFE 650 APPENDIX 6: FILTERS AND NETS Let (X, τ ) be a topological space, (D, ) a directed A6.3.5 Deﬁnition. set and φ : D → X a net in (X, τ ). Then φ is said to converge to a point a ∈ X, denoted by φ → a or {xα} → a or a ∈ lim{xα}, and a is said to be a limit of the net φ, if for each neighbourhood U in (X, τ ) of a, there exists a β ∈ D, such that φ(α) ∈ U , for every α ∈ D with β α. If in Deﬁnition A6.3.5 the limit of the net {xα} is unique, then we write lim{xα} = a. A6.3.6 Proposition. Hausdorﬀ if and only if no net in (X, τ ) converges to more than one point. Let (X, τ ) be a topological space. Then it is Proof. Exercise. Let (X, τ ) be a topological space and {xα} a net in A6.3.7 Deﬁnition. (X, τ ), where α is in a directed set (D, ). If Y is a subset of X, then {xα} is said to be eventually in Y if there exists a β ∈ D such that xα ∈ Y for all β α. Using Deﬁnition A6.3.7 we can rephrase Deﬁnition A6.3.5 as follows: Let (X, τ ) be a topological space, (D, ) a directed A6.3.8 Deﬁnition. set, {xα} a net in (X, τ ), α ∈ D, and a ∈ X. Then {xα} → a if for each neighbourhood U of a, xα is eventually in U . Let (X, τ ) be a topological space and {xα} a net in A6.3.9 Deﬁnition. (X, τ ), where α is in a directed set (D, ). If Y is a subset of X, then {xα} is said to be frequently or coﬁnally in Y if for each γ ∈ D, there exists a β ∈ D such that γ β and xβ ∈ Y At this point we discuss the equivalence of ﬁlterbases and nets. 651 Let (X, τ ) be a topological space, S a subset of X A6.3.10 Proposition. and a a point in X. Then a is a limit point of S if and only if there is a net in S converging to a. Proof. Firstly let a be a limit point of S. Deﬁne a directed set D by D = {U : U is a neighbourhood of a} where, for U1, U2 ∈ D, U1 U2 if U1 ⊇ U2. As a is a limit point of S, U ∩ S = Ø, for all U ∈ D. For each U ∈ D, let aU be an arbitary point of U ∩ S. Then {aU } is a net. If V is any neighbourhood of a, then V ∈ D. For every U V , aU ∈ U ⊆ V . So we have that {aU } → a. Conversely assume that a ∈ X and {xα} → a, where xα ∈ S, for α in a directed set D. Let U be any neighbourhood of a. Then there exists a β ∈ D, such that β γ =⇒ xγ ∈ U . So U ∩ S = Ø. Thus a is a limit point of S, as required. Let (X, τ ) be a topological space and S a subset of A6.3.11 Corollary. X. Then S is a closed set in (X, τ ) if and only if no net in S converges to a point of X \ S. Proof. Exercise. A6.3.12 Remark. Corollary A6.3.11 shows that nets (more precisely, convergent nets) can be used to describe the closed sets of (X, τ ), and therefore they also determine which subsets of (X, τ ) are open sets. In short, nets (more precisely, convergent nets) determine the topology τ on X. This result can be compared with Corollary 6.2.12 where it was shown that ﬁlterbases (more precisly, convergent ﬁlterbases) determine the topology τ on X. 652 APPENDIX 6: FILTERS AND NETS Let X be a set and {xα} a net in X, where α ∈ D, A6.3.13 Proposition. for a directed set D. For each β ∈ D, let Fβ = {xα : β α}. Then G = {Fβ : β ∈ D} is a ﬁlterbasis on X. Proof. Exercise. Let (X, τ ) be a topological space. A6.3.14 Deﬁnitions. (i) Let D be a directed set and {xα}, α ∈ D, a net in (X, τ ). For each β ∈ D, let Fβ = {xα : β α}. Put G = {Fβ : β ∈ D}. Then G is said to be the ﬁlterbase associated with the net {xα}, α ∈ D. If F is the ﬁlter generated by the ﬁlterbase G, then F is said to be the ﬁlter associated with the net {xα}, α ∈ D. (ii) Let G be a ﬁlterbase on (X, τ ) which generates the ﬁlter F . Deﬁne the set D = {(x, F ) : x ∈ F ∈ F}. Deﬁne a partial ordering on D as follows: (x1, F1) (x2, F2) if F2 ⊆ F1. Then D is a directed set. Deﬁne φ : D → X by φ((x, F )) = x. Then φ is said to be the net associated with the ﬁlter F and the net associated with the ﬁlterbase G. Let (X, τ ) be a topological space and a ∈ X. A6.3.15 Proposition. (i) Let {xα}, α ∈ D, D a directed set, be a net on (X, τ ). Let F be the ﬁlter (respectively, ﬁlterbase) associated with the net {xα}, α ∈ D. Then F → a if and only if {xα} → a. (ii) Let F be a ﬁlter (respectively, ﬁlterbase) on (X, τ ). Then the net {xα}, α ∈ D, associated with the ﬁlter (respectively, ﬁlterbase) F . Then F → a if and only if {xα} → a. Proof. Exercise. Let (X, τ ) be a topological space, D a directed set, A6.3.16 Deﬁnition. and {xα}, α ∈ D a net in (X, τ ). Then {xα}, α ∈ D is said to be an ultranet (or a universal net) if for every subset S of X this net is eventually in either S or X \ S; that is, there exists a β ∈ D such that either xα ∈ S for all β α or xα ∈ X \ S for all β α 653 A6.3.17 Proposition. set, {xα}, α ∈ D a net on (X, τ ), and F a ﬁlter on (X, τ ). (i) If {xα}, α ∈ D, is an ultranet and F is the ﬁlter associated with this net, Let (X, τ ) be a topological space, D a directed then F is an ultraﬁlter. (ii) If F is an ultraﬁlter and {xα}, α ∈ D, is the net associated with this ﬁlter, then {xα}, α ∈ D, is an ultranet. Proof. Exercise. Let X and Y be sets, D a directed set and {xα}, A6.3.18 Proposition. α ∈ D, a net in X. If f is any function of X into Y , then {f (xα)}, α ∈ D, is a net in Y . Proof. Exercise. Let (X, τ 1) and (Y, τ 2) be topological spaces and A6.3.19 Proposition. f a function of X into Y . The map f : (X, τ 1) → (Y, τ 2) is continuous if and only if for each point a in X and each net {xα} converging to a in (X, τ 1), the net {f (xα)} converges to f (a) in (Y, τ 2). Proof. Exercise. 654 APPENDIX 6: FILTERS AND NETS Now we deﬁne the notion of a subnet, which is a generalization of that of a subsequence. However, it is slightly more technical than one might at ﬁrst expect. This extra technicality is needed in order that certain properties of subnets that we want will be true. Let X be a set, D1 and D2 directed sets, and A6.3.20 Deﬁnitions. φ1 : D1 → X and φ2 : D2 → X nets in X. A function θ : D2 → D1 is said to be non-decreasing if β1 β2 =⇒ θ(β1) θ(β2), for all β1, β2 ∈ D2. The function θ is said to be coﬁnal if for each α ∈ D1, there exists a β ∈ D2 such that θ(β) α. The net φ2 is said to be a subnet of the net φ1 if there exists a non-decreasing coﬁnal function θ : D2 → D1 such that φ2 = φ1 ◦ θ. A6.3.21 Remark. In the literature there are two inequivalent deﬁnitions of subnet. As well as that used in Deﬁnitions A6.3.20, another deﬁnition does not include that θ is non-decreasing but rather only that it is coﬁnal. Let D1 = D2 = N with the usual ordering, let θ : N → N A6.3.22 Example. be given by θ(n) = 3n, for n ∈ N, and (X, τ ) any topological space. Clearly θ is a non-decreasing coﬁnal map. So if x1, x2, . . . , xn, . . . is any sequence in (X, τ ) then it is also a net with directed set N and the subsequence x3, x6, . . . , x3n, . . . is also a subnet. for example [ 5 Let D1 = D2 = N with the usual ordering, let θ : N → N A6.3.23 Example. be given by θ(n) = 1 + [n 2 ], n ∈ N, where [x] denotes the integer part of x, [5] = 5. The map θ is non-decreasing [4.9] = 4, in (X, τ ) has a subnet and coﬁnal. So the sequence and net x1, x2, . . . , xn, . . . 2 ], . . . which is not a subsequence. So while x1, x2, x2, x3, x3, x4, x4, x5, . . . , x1+[ n every subsequence of a sequence is a subnet, not every subnet of a sequence is 3] = 1, a subsequence. 655 Let τ be the discrete topology on the set X of all positive A6.3.24 Example. integers. Consider the directed sets D1 = N and D2 = (1, ∞), with the usual orderings. Let φ1 : N → X be the identity map and θ : (1, ∞) → N be given by θ(x) = [x], the integer part of x, x ∈ (1, ∞). Let φ2 : (1, ∞) → X be deﬁned by φ2 = φ1 ◦ θ. Clearly φ1 and φ2 are nets in X and φ2 is a subnet of φ1 by its very deﬁnition. It is interesting to note that the directed set D1 = N is a countable set, while the directed set D2 = (1, ∞) is an uncountable set. In other words, if we put xn = φ1(n), n ∈ N and xr = φ2(r), for r ∈ (1, ∞), then the uncountable net {xr}, r ∈ (1, ∞), is a subnet of the countable sequence (and net) {xn}, n ∈ N. Let (X, τ ) be a topological space and D a directed A6.3.25 Deﬁnition. set. The point a is said to be a cluster point of the net {xα}, α ∈ D, if for each neighbourhood U of a and each β ∈ D, there exists an α ∈ D such that α β and xα ∈ U . Let (X, τ ) be a topological space, D a directed A6.3.26 Proposition. set, a ∈ X, and {xα}, α ∈ D a net in (x, τ ). If {xα} → a then a is a cluster point of the net {xα}. Proof. Exercise. The next proposition tells us that a cluster point of a subnet of a net is also a cluster point of the net. 656 APPENDIX 6: FILTERS AND NETS Let (X, τ ) be a topological space, D1 and D2 A6.3.27 Proposition. directed sets, a ∈ X, φ : D1 → X a net in (X, τ ) with subnet φ ◦ θ : D2 → X, where θ : D2 → D1 is a non-decreasing coﬁnal map. If a is a cluster point of the subnet φ ◦ θ : D2 → X then a is also a cluster point of the net φ : D1 → X. Proof. Exercise. Let (X, τ ) be a topological space, D1 a directed A6.3.28 Proposition. set, a ∈ X, and {xα}, α ∈ D1 a net in (X, τ ). Then a is a cluster point of the net {xα}, α ∈ D1, if and only if the net has a subnet which converges to a. Firstly assume that the net {xα}, α ∈ D1, has a as a cluster point. Proof. Put D2 = {(α, U ) : α ∈ D1, U a neighbourhood of a such that xα ∈ U }. Deﬁne a partial ordering on D2 by (α1, U1) (α2, U2) when α1 α2 and U1 ⊇ U2. We claim that D2 is a directed set. To see this let (α1, U1) ∈ D2 and (α2, U2) ∈ D2. As α1, α2 ∈ D1 and D1 is a directed set, there exists an α3 ∈ D1 with α1 α3 and α2 α3. Put U3 = U1 ∩ U2, so that we have U1 ⊇ U3 and U2 ⊇ U3. As a is a cluster point and a ∈ U
3, there exist an α4 with α3 α4 such that xα4 ∈ U3. So (α4, U3) ∈ D2. As α1 α4, α2 α4, U1 ⊇ U3 and U2 ⊇ U3, we see that (α1, U1) (α4, U3) and (α2, U2) (α4, U3). So D2 is indeed a directed set. Deﬁne a map θ : D2 → D1 by θ(α, U ) = α, where α ∈ D1 and U is a neighbourhood of a. Clearly θ is non-decreasing. Now consider any α ∈ D1. As a is a cluster point, for any neighbourhood U of a, there exists a γ ∈ D1 with α γ such that xγ ∈ U . So (γ, U ) ∈ D2 and we have θ((γ, U )) = γ and α γ. So θ is also coﬁnal. For notational convenience, write φ : D1 → X where φ(α) = xα, α ∈ D1. Then φ ◦ θ : D2 → X is a subnet of the net {xα}, α ∈ D1. As a is a cluster point of the net {xα}, α ∈ D1, for each neighbourhood U0 of a and each α0 ∈ D1, there exists a β ∈ D1, α0 β such that xβ ∈ U0. Then by the deﬁnition of D2, for any (α, U ) ∈ D2 with (β, U0) (α, U ) we have xα ∈ U and In other words, φ ◦ θ(α, U ) ⊆ U0, for all (β, U0) (α, U ) ; U ⊆ U0. So xα ∈ U0. that is the subnet φ ◦ θ converges to a. The converse follows immediately from Propositions A6.3.26 and A6.3.27. A6.3.29 Proposition. compact if and only if every ultranet in (X, τ ) converges. Let (X, τ ) be a topological space. Then (X, τ ) is 657 Proof. This follows immediately from Propositions A6.1.24, A6.3.15 and A6.3.17. A6.3.30 Proposition. Every net has a subnet which is an ultranet. Proof. Exercise. A6.3.31 Proposition. Let (X, τ ) be a topological space. The following are equivalent; (i) (X, τ ) is compact; (ii) Every net in (X, τ ) has a subnet which converges; (iii) Every net in (X, τ ) has a cluster point. Proof. Exercise. Exercises A6.3 1. Verify that the partially ordered sets in Examples 10.2.3 and Examples 10.2.4 are directed sets. 2. Verify that if (X, τ ) is a Hausdorﬀ space, then a net in (X, τ ) converges to at most one point. 3. Prove the converse of the result in Exercise 2 above, namely that if (X, τ ) is a topological space such that no net converges to two distinct points of X, then (X, τ ) is Hausdorﬀ. 658 APPENDIX 6: FILTERS AND NETS 4. (i) Let S be any inﬁnite set and, using the Axiom of Choice, select subsets S1, S2, . . . , Sn, . . . , n ∈ N, such that S = ∞ n=1 Sn. (ii) Using the Well-Ordering Theorem on each Sn, n ∈ N , deﬁne a partial ordering on all of S such that x y if x ∈ Si and y ∈ Sj, where i j. (iii) Verify that with this ordering, S is a directed set. (iv) Deduce that for every inﬁnite cardinal number, ℵ, there is a directed set of cardinality ℵ. 5. Prove Corollary A6.3.11. 6. Prove Proposition A6.3.13. 7. Prove Proposition A6.3.15. 8. Prove Proposition A6.3.17. 9. Prove Proposition A6.3.18. 10. Prove Proposition A6.3.19. 11. Let D1 = D2 = R with the usual ordering. Which of the following maps θ : D2 → D1 are non-decreasing and which are coﬁnal? (i) θ(x) = x2, for all x ∈ R. (ii) θ(x) = \|x\|, for all x ∈ R. (iii) θ(x) = sin x, for all x ∈ R. (iv) θ(x) = 2x, for all x ∈ R. (v) θ(x) = 2x3 − 3x, for all x ∈ R. 12. Prove the following: (i) Every net is a subnet of itself. (ii) If {xγ}, γ ∈ D3, is a subnet of the net {xβ}, β ∈ D2, and {xβ}, β ∈ D2, is a subnet of the net {xα}, α ∈ D1, then {xγ}, γ ∈ D3, is also a subnet of {xα}, α ∈ D1. 13. Let (X, τ ) be any topological space, where X is an inﬁnite set. Let D1 = D2 = Q and x1, x2, x3, . . . , xn, . . . a sequence in (X, τ ). Find two subnets of this sequence neither of which is a subsequence of the sequence x1, x2, x3, . . . , xn, . . . . 14. Using Exercise 4 above and Example A6.3.24, prove that for each cardinal number ℵ, every inﬁnite sequence x1, x2, . . . , xn, . . . (X, τ ) has a subnet {xα}, α ∈ D, where D has cardinality ℵ. in a topological space 659 15. Prove Proposition A6.3.26. 16. Prove Proposition A6.3.27. 17. Prove Proposition A6.3.30. 18. Using Propositions A6.3.29, A6.3.30, and A6.3.28 prove Proposition A6.3.31. A6.4 Wallman Compactiﬁcations: An Application of Ultraﬁlters Repeatedly throughout the study of topology we are interested in whether a given topological space can be embedded as a subspace of a topological space with nicer properties. In metric space theory we know that not every metric space is complete. But can every metric space be embedded as a metric subspace of a complete metric space? The answer is “yes”, and was proved in Proposition 6.3.23. This was achieved by ﬁrst recognizing that a metric space is complete if and only if every Cauchy sequence converges. So a metric space which fails to be complete must have Cauchy sequences which are not convergent. For example in Q with the euclidean metric, we easily ﬁnd Cauchy sequences of rational numbers which do not converge (to any rational number). So we need to enlarge the metric space with extra points so that the Cauchy sequences which did not converge previously now converge to one of the extra points. In the case of Q, we add what is eﬀectively the irrational numbers, and the completion is R. So the “trick” or rather the “technique”, is to identify why the given topological space fails to have the desired property and then add extra points in such a way that the topological space with the extra points does have the desired property. 660 APPENDIX 6: FILTERS AND NETS In 1937 M.H. Stone and E. ˘Cech introduced what is now known as the Stone˘Cech compactiﬁcation which is deﬁned in Deﬁnition 10.4.1. They showed that a topological space can be embedded as a subspace of a compact Hausdorﬀ space if and only if it is completely regular and Hausdorﬀ, that is it is a Tychonoﬀ space. This is discussed in §10.4. In 1938 H. Wallman (Wallmani [418]) proved that a topological space can be embedded as a subspace of a compact T1-space if and only of it is a T1-space. We shall now describe the Wallman compactiﬁcation. If we are given a T1-space which is not compact, how can we enlarge the space to obtain a compact T1-space? To answer this we ﬁnd a convenient property for testing compactness. We know several such properties. We look for one which is most convenient. As indicated in the title of this section, we shall use ultraﬁlters. Note Proposition 10.3.2 states that a topological space (X, τ ) is compact if and only if for every family F of closed subsets of X with the ﬁnite intersection (F.I.P.) property (see Deﬁnition 10.3.1) F = Ø. F ∈F So we shall focus our intention on families of closed sets with the ﬁnite intersection property. Now we know that ﬁlters and ultraﬁlters, in particular, have the F.I.P. But, in almost all topological spaces, families of closed sets cannot form a ﬁlter, since any set which contains a set in the ﬁlter is also in the ﬁlter. Therefore we shall modify the notion of an ultraﬁlter so that the modiﬁcation still has all the desired properties of an ultraﬁlter but can consist of only closed sets. Let (X, τ ) be a topological space and C the set of all A6.4.1 Deﬁnitions. closed subsets of X. A non-empty subset F of C is said to be a ﬁlter in C if (i) F1, F2 ∈ F implies F1 ∩ F2 ∈ F ; (ii) F ∈ F and F ⊆ G ∈ C =⇒ G ∈ F ; and (iii) Ø /∈ F . A ﬁlter U in C is said to be an ultraﬁlter in C if no ﬁlter in C is strictly ﬁner than U . An ultraﬁlter U in C is said to be a free ultraﬁlter in C if U ∈U U = Ø. We record some useful facts about ultraﬁlters in the set of all closed sets, C, on a topological space (X, τ ). The proof of Proposition A6.4.2 is analagous to that of The Ultraﬁlter Lemma A6.1.10 and Corollary A6.1.12. 661 Let F be any ﬁlter in C, the set of all closed sets, A6.4.2 Proposition. on a topological space (X, τ ). Then there exists an ultraﬁlter U in C which is ﬁner than F . Indeed if S is a non-empty set of closed subsets of X and S has the F.I.P., then there is an ultraﬁlter U in C such that S ⊆ U . 662 APPENDIX 6: FILTERS AND NETS Let U be an ultraﬁlter in C, the set of closed sets in A6.4.3 Proposition. a topological space (X, τ ). (i) F ∈ U and F ⊆ G ∈ C =⇒ G ∈ U ; (ii) Let F1, F2 ∈ C. Then F1, F2 ∈ U ⇐⇒ F1 ∩ F2 ∈ U ; (iii) Ø /∈ U ; (iv) F1, F2 ∈ U =⇒ F1 ∩ F2 = Ø; (v) {F : F ∈ U} has the F.I.P.; (vi) Let A, B ∈ C. Then A ∪ B ∈ U ⇐⇒ A ∈ U or B ∈ U ; (vii) Let A ∈ C. Then A ∈ U ⇐⇒ A ∩ F = Ø, for all F ∈ U ; (viii) Let U and U be ultraﬁlters in C. Then U = U ⇐⇒ there exist sets A ∈ U and B ∈ U such that A ∩ B = Ø; (ix) Let A ∈ C. Then A /∈ U ⇐⇒ A ⊆ X \ F, for some F ∈ U ; (x) Let O ∈ τ . Then X \ O /∈ U ⇐⇒ F ⊆ O, for some F ∈ U. Proof. (i), (ii), and (iii) follow from Deﬁnitions A6.4.1 (i), (ii) and (iii), respectively, while (iv) is a consequence of (1) and (iii) and (v) is a consequence of (iv). In (vi) =⇒ is proved analogously to Proposition A6.1.13. The converse follows from (i) above. In (vii) =⇒ follows from (iv). The proof of the converse in (vii) is analogous to the second paragraph of the proof of Proposition A6.1.13. (viii) follows from (vii) and (iv). (ix) follows from (vii). (x) follows from (ix). If U is any ultraﬁlter in C, then it has the ﬁnite interesection property. If U is a free ultraﬁlter, then (X, τ ) is not compact by Proposition 10.3.2. Let F be the set of all free ultraﬁlters in C. We shall expand the space (X, τ ) by adding extra “points” using these free ultraﬁlters. 663 Let ωX be the set X ∪ F. So each point in ωX is either a point in X or a free ultraﬁlter A6.4.4 Remarks. in C. At ﬁrst sight this might appear strange, but a set is just a collection of objects. In this case there are two kinds of objects in ωX, namely points and free ultraﬁlters in C. Our task is to deﬁne a topology on ωX in such a way that it has (X, τ ) as a subspace, and that with this topology ωX is a compact T1-space. For each open subset O of (X, τ ), deﬁne the subset O∗ of ωX by O∗ = O ∪ {U ∈ F : X \ O /∈ U}. (1) Then X∗ = X ∪ {U ∈ F : Ø /∈ U} = X ∪ F = ωX and O∗ = Ø ⇐⇒ O = Ø. (2) For each closed subset A of (X, τ ), deﬁne the subset A∗ of ωX by A∗ = A ∪ {U ∈ F : A ∈ U}. (3) Then X∗ = X ∪ {U ∈ F : X ∈ U} = X ∪ F = ωX and A∗ = Ø ⇐⇒ A = Ø. (4) Noting that (X, τ ) is a T1-space, for x ∈ X, {x} is a closed set in (X, τ ) and so {x}∗ = {x} ∪ {U ∈ F : {x} ∈ U} = {x} ∪ Ø = {x} (5) We shall show t
hat B = {O∗ : O ∈ τ } is a basis for a topology τ ω on ωX and that this topological space (ωX, τ ω) is a compact T1-space which has (X, τ ) as a subspace. 664 APPENDIX 6: FILTERS AND NETS We claim that for any O ∈ τ O∗ = ωX \ (X \ O)∗ . Proof of (6). ωX \ (X \ O)∗ = (X ∪ F) \ ((X \ O) ∪ {U ∈ F : X \ O ∈ U}), by (3) = (X \ (X \ O)) ∪ (F \ {U ∈ F : X \ O ∈ U}) = O ∪ {U ∈ F : X \ O /∈ U} = O∗ , by (1), which completes the proof. Similary for any closed set A in (X, τ ), we can show that A∗ = ωX \ (X \ A)∗ . We claim that for any A1, A2 ∈ C, (A1 ∩ A2)∗ = A1∗ ∩ A2∗ . Proof of (8). (6) (7) (8) A1∗ ∩ A2∗ = (A1 ∪ {U ∈ F : A1 ∈ U}) ∩ (A2 ∪ {U ∈ F : A2 ∈ U}), by (3) = (A1 ∩ A2) ∪ {U ∈ F : A1, A2 ∈ U} = (A1 ∩ A2) ∪ {U ∈ F : A1 ∩ A2 ∈ U}, by Proposition A6.4.3(ii) = (A1 ∩ A2)∗ , by (3), which completes the proof Similarly we can show that (A1 ∪ A2)∗ = A1∗ ∪ A2∗ . We claim that for O1, O2 ∈ τ (O1 ∪ O2)∗ = O1 ∗ ∪ O2 ∗ . (10) follows from (6) and (3) and so its proof is left as an exercise. We also leave the proof of (11) as an exercise. (O1 ∩ O2)∗ = O1 ∗ ∩ O2 ∗ . (9) (10) (11) 665 As a consequence of (8) and the second part of (4) we obtain Let A1, A2, . . . , An ∈ C. Then n i=1 Ai = Ø ⇐⇒ n i=1 Ai∗ = Ø. (12) It follows immediately from (12) that for Ai ∈ C, i ∈ I, any index set, {Ai : i ∈ I} has the F.I.P. ⇐⇒ {Ai∗ : i ∈ I} has the F.I.P.. (13) Now by Proposition 2.2.8, (11) and (2) imply that the set B = {O∗ : O ∈ τ } is indeed a basis for a topology τ ω on ωX. So every open set in (ωX, τ ω) is a union of members of B. Every closed set in (ωX, τ ω) is therefore an intersection of closed sets A∗, where A is closed in (X, τ ). A6.4.5 Deﬁnition. (ωX, τ ω) is called the Wallman compactiﬁcation of X. Let (X, τ ) be a T1-space.Then the topological space 666 APPENDIX 6: FILTERS AND NETS Let (X, τ ) be a T1-space. Then its Wallman A6.4.6 Theorem. compactiﬁcation (ωX, τ ω) is a compact T1-space that contains (X, τ ) as a dense subspace. Further, every continuous map of (X, τ ) into a compact Hausdorﬀ space (K, τ 1) extends to a continuous map of (ωX, τ ω) into (K, τ 1). Proof. The fact that (X, τ ) is a subspace of (ωX, τ ω) follows immediately from the deﬁnitions of ωX, B, τ ω, and O∗, for O ∈ τ . That X is dense in (ωX, τ ω) follows immediately from the the deﬁnition of the basic open sets O∗, O ∈ τ , as each (non-empty) O∗ intersects X non-trivially. Next we shall prove that (ωX, τ ω) is a T1-space, that is each point is a closed set. We know that there are two kinds of points in ωX, namely x ∈ X and U , where U ∈ F. By (4) above each {x} = {x}∗ and so is a closed set in (ωX, τ ω). Noting that if U, F ∈ F are distinct ultraﬁlters, there exists A ∈ U such that A /∈ F , because otherwise F would be a ﬁner ﬁlter in C than U , which contradicts U being an ultraﬁlter in C. Therefore {F ∈ F : A ∈ F} = {U}. (14) A∈U Using (14), the fact that U is a free ultraﬁlter in C, and (3) which says that A∗ = A ∪ {F ∈ F : A ∈ F} we obtain that  A∗ =  A∈U A∈U  A    A∈U {F ∈ F : A ∈ F}   = (Ø) ∪ ({U}) = {U}. Thus U is the intersection of the closed sets A∗, A ∈ U , and so is a closed set. So we have that every {x} and every U is a closed set in (ωX, τ ω). Thus (ωX, τ ω) is indeed a T1-space. Finally, we need to show that (ωX, τ ω) is compact. Let I be an index set and {Ci : i ∈ I} a set of closed subsets of ωX with the F.I.P. We are required to prove that i∈I As noted earlier, each Ci = j∈J , for some index set J , where each Aij Ci = Ø. (Aij)∗ is a closed subset of X. Therefore {(Aij)∗ : i ∈ I, j ∈ J} has the F.I.P.. By (12) If U ∈U x ∈ i∈I j∈J If U ∈U U ∈ (Aij)∗ this implies that {Aij : i ∈ I, j ∈ J} has the F.I.P.. So by Proposition A6.4.2 there exists an ultraﬁlter U in C containing {Aij : i ∈ I, j ∈ J}. 667 U = Ø, then there exists an x ∈ X with x ∈ U ∈U Aij and so x ∈ i∈I . Thus i∈I (Aij)∗ j∈J Ci = Ø, as required. U . This implies j∈J (Aij)∗ = Ci = Ø, as required. U = Ø, U is a free ultraﬁlter in C. Now each Aij ∈ U and so, by (2), . Therefore U ∈ i∈I Ci. Thus i∈I So in both cases, i∈I i∈I Ci = Ø. Hence (ωX, τ ω) is compact. Let φ : (X, τ ) be a continuous map of (X, τ ) into (K, τ 1). We shall apply Proposition 10.3.53 to obtain the required result. So let C1, C2 be disjoint closed subsets of (K, τ 1). As φ : (X, τ ) → (K, τ 1) is continuous, φ−1(C1) and φ−1(C2) are disjoint closed subsets of (X, τ ). So by (8) and (2), [φ−1(C1)]∗ and [φ−1(C2)]∗ are disjoint closed subsets of (ωX, τ ω). As [φ−1(C1)]∗ is a closed set containing φ−1(C1) and [φ−1(C1)]∗ is a closed set containing φ−1(C2), the closures in ωX of φ−1(C1) and φ−1(C2) are disjoint. By Proposition 10.3.53, φ : (X, τ ) → (K, τ 1) has a continuous extension (ωX, τ ω) → (K, τ 1), which completes the proof of the theorem. 668 APPENDIX 6: FILTERS AND NETS A6.4.7 Proposition. Let (X, τ ) be a T1-space. The following conditions are equivalent: (i) The Wallman compactiﬁcation (ωX, τ ω) is a Hausdorﬀ space; (ii) (X, τ ) is a normal space. (i) =⇒ (ii): If (ωX, τ ω) is a Hausdorﬀ space, it is compact Hausdorﬀ Proof. which, by Remark 10.3.28, implies that it is a normal space. Now let C1 and C2 be disjoint closed subsets of (X, τ ). Then by Remarks A6.4.4 (8) and (2), (C1)∗ and (C2)∗ are disjoint closed sets in the normal space (ωX, τ ω). So there exist disjoint O1, O2 ∈ τ ω, such that (C1)∗ ⊆ O1 and (C2)∗ ⊆ O2. Then the disjoint sets O1 ∩ X and O2 ∩X are open sets in (X, τ ) which respectively contain C1 and C2. So (X, τ ) is a normal space. (ii) =⇒ (i): Assume (X, τ ) is normal. We need to show that if z1, z2 ∈ ωX = X ∪ F , then there are disjoint open sets containing z1 and z2 respectively. So we need to consider the cases: (a) z1, z2 ∈ F ; (b) z1 ∈ X and z2 ∈ F ; (c) z1, z2 ∈ X. (a): Let U1, U2 ∈ F be distinct. By Proposition A6.4.3 (viii), there exist closed subsets A1 and A2 of (X, τ ) such that A1 ∈ U1 and A2 ∈ U2 and A1 ∩ A2 = Ø. As (X, τ ) in normal, there exist disjoint open sets U1, U2 in (X, τ ) such that A1 ⊆ U1 2 are disjoint open sets in (ωX, τ ω). By Proposition and A2 ⊆ U2. By (11), U ∗ A6.4.3 (x), this implies that X \ U1 /∈ U1 and X \ U2 /∈ U2. By (1) these imply U1 ⊆ U ∗ 2 , which complete the proof for case (a). 1 and U2 ⊆ U ∗ 1 and U ∗ (b): Let x1 ∈ X and U2 ∈ F . As U2 is a free ultraﬁler in C, there exists A2 ∈ U2 with x1 ∈ A2. Put A1 = {x1}. As in (a), there exist disjoint open sets U1, U2 in (X, τ ) containing A1 and A2 respectively, such that U ∗ 2 are disjoint open sets in (ωX, τ ω) with x1 ∈ U1 and U2 ∈ U2. This completes the proof of case (b). (c): Let x1, x2 ∈ X. Then there exist disjoint open sets U1, U2 in (X, τ ) which 2 are disjoint open sets in (ωX, τ ω) 1 and U ∗ contain x1 and x2, respectively. Then U ∗ which contain x1 and x2, respectively. This completes the proof of case (c) and of the proposition. 1 and U ∗ A6.4.8 Corollary. (i) (X, τ ) is a normal Hausdorﬀ space. (ii) The Wallman compactiﬁcation ωX is the Stone-˘Cech compactiﬁcation βX. The following conditions are equivalent: 669 Proof. Exercise. A6.4.9 Corollary. space and τ be the subspace topology on X. Then Let X be any unbounded subset of a normed vector βX = ωX and card (βX) = card (ωX) 2c. In particular, for non-negative integers a, b, c, and d with a + b + c + d > 0, card (ω(Na × Qb × Pc × Rd)) = 2c. Proof. Exercise. A6.4.10 Corollary. then card (ωX) = 22m . If (X, τ ) is a discrete space of inﬁnite cardinality m, Proof. Exercise. A6.4.11 Proposition. 22m distinct ultraﬁlters. If X is any inﬁnite set of cardinality m, then it has Proof. Put the discrete topology on the set X. Then ωX = X ∪ F , where F is the set of free ultraﬁlters. (On a discrete space every free ultraﬁlter is a free ultraﬁlter on the closed sets.) 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Index [0, 1], 263, 264, 274, 275, 300, 301, 327, almost-periodic functions, 599 332, 335 ωω-base, 313 ℵ0, 394 abelain, 526 abelian group, 493 torsion-free, 532 absolutely convex, 209 absolutely convex set, 209 absorbent set, 209 AC, 632 accumulation point, 73 Acta Arithmetica, 434 adjoint, 388 Alaoglu, Leonidas, 316 Alexander Subbasis Theorem, 179 algebra C∗-, 388 associative, 376 Banach, 377 analytic set, 433 analytic function, 366 analytic set, 154 annihilator, 320, 601, 610 antisymmetric binary relation, 265, 378 arrow, 110 Ascoli’s Theorem, 554 Asplund space, 324 Asplund, Edgar, 324 associative algebra, 376 attracting ﬁxed point, 442 attracting periodic point, 452 axiom least upper bound, 85 Axiom of Choice, 269, 631, 632 Bn, 344 Bézier curves, 364 Bézierr, Pierre, 364 commutative, 376 Baire Category Theorem, 164, 165 normed, 377 unital, 376 for Locally Compact Spaces, 534 Baire space, 165 algebra over a ﬁeld, 376 Baire, René Louis, 426 algebraic number, 398 balanced, 210 almost connected, 601, 607 ball 704 INDEX closed unit, 187 Banach Fixed Point Theorem, 162 Banach algebra, 377 bijection, 393 bijective, 41 bikompakt, 192 binary relation 705 Banach space, 158, 265, 525 antisymmetric, 265, 378 compactly generated, 208 reﬂexive, 97, 265, 378 reﬂexive, 315, 321, 322 separable, 159, 208, 319 symmetric, 97 transitive, 97, 265, 378 separable dual, 319 Birkhoﬀ weakly compactly generated, 322 George David, 462 Banach space dual reﬂexive, 321 Bohr compactiﬁcation, 599 Bolzano-Weierstrass Theorem, 151 Banach, Stefan, 316, 317, 427 bound Banach-Alaoglu Theorem, 316, 322 greatest lower, 85, 379 Banach-Mazur Theorem, 386 Banach-Tarski paradox, 427 least upper, 379 lower, 85, 379 Banachiewicz, Tadeusz, 432 upper, 85, 268, 269, 379 base ﬁlter, 640 ultraﬁlter, 648 basis, 57, 544 ﬁlter, 640 ultraﬁlter, 648 Battista, Leon, 488 Bernoulli,Jacob, 354 boundary, 164 boundary point, 164 bounded, 85, 183 above, 85 below, 85 metric, 134 metric space, 134 set, 160 Bernstein base polynomials, 308 bounded linear functional, 316 Bernstein polynomial, 309, 363 bounded linear operator, 316 Besicovitch bounded subset of a poset, 312 Abram Samoilovitch, 470 box topology, 225 bifurcation, 452 Brouwer Fixed Point Theorem, 120 period doubling, 453 Brouwer, Luitzen Egbertus Jan, 428 Bifurcation Theorem Brunelleschi, Filippo, 488 First, 452 bump function, 370 706 butterﬂy eﬀect, 461 (C(X), F ), 370 (C(X), C), 370 (C(X), R), 370 C[0, 1], 68, 126, 127, 187, 365 C∗-algebra, 388 c, 406 c0, 136, 159 C∞, 367 CAD, 364 Café, Scottish, 427 Cantor Georg, 401 Cantor set, 222 Cantor Space middle two-thirds, 248 Cantor space, 222 Cantor, Georg, 432 card , 406 cardinal function, 312 cardinal invariant, 312 cardinal number, 406 cardinality, 394 carpet Sierpiński, 433 Cartan, Élie, 506 INDEX manifold, 140 cellularity, 312 centre, 526 cf(P ), 312 chaos, 455, 470 chaotic dynamical system, 463 character, 556 character group, 556 Church-Kleene ordinal, 423 circle group, 493 Clebsch, Rudolf Friedrich Alfred, 485 clopen set, 36 closed mapping, 186, 198, 340 sequentially, 146 set, 34 unit ball, 187 closure, 77 cluster point, 73 cluster point of ﬁlterbase, 643 cluster point of net, 655 coarser, 622 coarser ﬁlterbasis, 641 coarser topology, 116, 198 coarsest topology, 261 coﬁnal, 312, 654 cartesian product, 260, 300 coﬁnality of a poset, 312 category ﬁrst, 166 second, 166 coﬁnally, 650 coﬁnite topology, 39 coinitial, 312 Cauchy sequence, 148 collectionwise Hausdorﬀ, 245 Cauchy-Riemann commutative algebra, 376 INDEX 707 commutator, 510 Compositio Mathematica, 430, 434 commutator group, 510 cone, 343 commutator subgroup, 526, 533 reduced, 347 compact, 176, 177, 265, 272–274, 277, conjugate dynamical systems, 466 279, 284, 299, 326, 631, 646 conjugate map, 466 countably, 189, 190 connected, 86, 298, 301 relatively, 188 sequentially, 190 space, 177 weak*-, 316 compact element, 608 compact Lie group, 496 compact space, 646 almost, 601, 607 component, 213 locally, 216, 252 manifold, 140 path, 118 pathwise, 118 connected topological group, 522 compact-open topology, 552 continuous, 110 compactiﬁcation Bohr, 599 Stone-˘Cech, 326, 329–332, 335 Wallman, 665 at a point, 170 homomorphism, 519 jointly, 552 sequentially, 147 compactly generated, 522, 582 continuous homomorphism, 519 compactly generated Banach space, 208 continuous mapping, 112 compactly-generated space, 303 continuum, 215 compactum, 215 comparable textbf, 268 complete, 149 contraction mapping, 161 Contraction Mapping Theorem, 162 contradiction, 53 completely metrizable, 153 converge, 141, 190, 642, 650 completely regular, 274, 276, 300 converge, 628 completely regular space, 275 convergent completion pointwise, 365 of a metric space, 155 convergent ﬁlterbase, 642 complex trignometric polynomial, 385 convergent sequence, 190 component, 121, 213, 215, 523, 526 Converse of Heine-Borel Theorem, 183 identity, 508 convex 708 set, 167 absolutely, 209 convex set, 209 Coolidge, Julian Lowel, 485 coordinates homogeneous, 491 cosmic space, 247 countability ﬁrst axiom of, 137 second axiom of, 62 countable second, 241 countable bases, 137 countable chain condition, 244 countable closed topology, 45 countable ordinaltextbf, 423 countable set, 394 countable tightness, 147 countable-closed space, 147 countably compact, 189, 190 countably inﬁnite, 394 Courant, Richard, 485 cover, 177 covering, 177 open, 177 CR-manifold, 140 INDEX curves Bézier, 364 cycle m, 452 cylinder, 206 de Casteljau, Paul, 364 decreasing sequence, 150 degree of trignometric polynomial, 384, 385 Delian problem, 486 Della pittura, 488 dense, 78 everywhere, 78 nowhere, 165 density, 312 density character, 288, 386 denumerable, 394 derivative partial, 368 derivative of a function of several variables, 368 derived set, 170 descriptive geometry, 490 Devaney,Robert L., 462 diagonal, 196, 550 cube, 274, 274, 276, 277, 279, 300 diam, 470 n, 230 Hilbert, 230, 263, 264, 274, 283, 285, diameter, 470 Dickstein, Samuel, 434 299 curve Sierpiński, 433 space-ﬁlling, 253, 433 Dieudonné, Jean Alexandre Eugène, 309, 497 diﬀerentiable, 163 manifold, 140 INDEX 709 diﬀerentiable at a point, 367, 368 double, 315 diﬀerentiable manifold, 367, 496 dual group, 502, 556 dimH , 473 dimension Hausdorﬀ, 473 projective, 490 zero, 122 Dini’s Theorem, 370 Dini, Ulisse, 370 diophantine, 549 direct product restricted, 529, 576 weak, 529 directected, 313 directed set, 649 disconnected, 87, 118 extremally, 637 totally, 121, 524, 526 discrete metric, 125 space, 26 topology, 26 distance, 124 distance between sets, 146 divisible somewhat, 613 dual space, 314 dyadic, 278 dynamical system, 441, 463 chaotic, 463 dynamical systems conjugate, 466 En, 344 ε-covering, 471 ε-number, 423 ε1, 423 ecology, 444 Egorov, Dimitri Feddrovich, 433 element greatest, 85 least, 85 embedded, 263, 277, 279, 283, 299 embedding, 265 isometric, 155 Embedding Lemma, 238, 262, 274, 277 empty union, 30 equicontinuous, 554 equicontinuous at a point, 554 equipotent, 393 divisible group, 207, 503, 530 equivalence class, 342 door space, 47, 94 double dual, 315 doubling function, 444 doubling the cube, 486 Dowker space, 307 dual equivalence relation, 109, 157, 342, 393 equivalent ﬁlterbases, 641 equivalent metric, 132 ergodic theory, 462 Erlangen program, 491 euclidean 710 INDEX locally, 140, 509 ﬁlter determined by the sequence, 641 euclidean geometry, 492 euclidean group, 495 euclidean metric, 125 euclidean metric on R2, 125 Euclidean topology, 621 euclidean topology, 51 euclidean topology on Rn, 60 evaluation map, 238, 262, 277, 327, 329 eventually, 650 eventually ﬁxed, 440 eventually periodic, 440 everywhere dense, 78 exact sequence, 590 exterior, 164 exterior point, 164 ﬁlter generated by a ﬁlterbasis, 640 ﬁlter in C, 660 ﬁlterbase, 640 limit point, 642 neighbourhood, 641 ﬁlterbase associated with net, 652 ﬁlterbases equivalent, 641 ﬁlterbasis, 640 coarser, 641 ﬁner, 641 ﬁnal segment topology, 31 ﬁner, 622 ﬁner ﬁlterbasis, 641 ﬁner topology, 116, 198 extremally disconnected, 637 extremally disconnected space, 245 ﬁnite, 394 locally, 306 Fσ-set, 55, 170 f −1, 42 F.I.P., 271 ﬁeld topological, 598 ﬁlter, 620 generated by S, 622 Fréchet, 621 free, 621 neighbourhood, 621 principal, generated by §, 621 ﬁlter associated with the net, 652 ﬁlter base, 640 ﬁlter basis, 640 ﬁnite intersection property, 189, 271, 272, 273, 622, 625, 661 ﬁnite space, 47 ﬁnite subcovering, 177 ﬁnite topological space, 47 ﬁnite-closed topology, 39 ﬁrst axiom of countability, 137, 300, 619 First Bifurcation Theorem, 452 ﬁrst category, 166 ﬁrst countable, 137, 300 ﬁrst uncountable ordinal, 423 ﬁxed point, 120, 161, 438 attracting, 442 eventually, 440 neutral, 442 INDEX repelling, 442 ﬁxed point property, 120 Fixed Point Theorem, 120 Banach, 162 Fréchet ﬁlter, 621 Fréchet, Maurice, 430 fractal, 433 geometry, 470 Frechet space, 147 Frechet topological vector space, 255 Frechet-Urysohn space, 147 free abelian group, 544 free abelian group basis, 544 free ﬁlter, 621 free ultraﬁlter in C, 660 frequently, 650 Freyd Adjoint Functor Theorem, 326 function analytic, 366 bijective, 41 bump, 370 continuous, 110 doubling, 444 injective, 41 inverse, 41 logistic, 444 one-to-one, 41 onto, 41 smooth, 366 surjective, 41 tent, 468 711 linear, 314 funct
ions almost-periodic, 599 Fundamenta Mathematica, 433 Fundamental Theorem of Algebra, 217 GL(n, C), 513 Gδ-set, 55, 170 ΓX, 347 G-base, 313 Galileo Galilei, 395 Gauss, Carl Friedrich, 485 Gelfand-Naimark Representation Theorem, 388 general linear group, 494, 513 Generalized Heine-Borel Theorem, 184, 204 geometry descriptive , 490 euclidean, 492 projective, 490 Georg Cantor, 401 graphical analysis, 444 greatest element, 85, 268, 269, 270 greatest lower bound, 85, 379 group, 493 abelian, 493 character, 556 circle, 493 commutator, 510 compact Lie, 496 functional divisible, 207, 503, 530 bounded linear, 316 dual, 502, 556 712 INDEX euclidean, 495 free abelian, 544 general linear, 494, 513 Heisenberg, 497 LCA, 523 Lie, 496 linear, 513 Lorentz, 492, 497 matrix, 494 Hahn-Banach Theorem, 318, 321 Hausdorﬀ s-dimensional outer measure, 472 collectionwise, 245 dimension, 473 Felix, 470 Hausdorﬀ space, 94, 133, 196, 285, 301, 515 Hausdorﬀ topological group, 527 Moskowitz-Morris, 605 Hausdorﬀ, Felix, 431 NSS-, 509 orthogonal, 494, 513 pro-Lie, 507 product, 494 proﬁnite, 506 projective, 495 quotient, 494, 523, 525 simple, 600 simply connected Lie, 506 special linear, 513 special orthogonal, 513 special unitary, 513 topological, 495, 512 topologically simple, 600 topology, 514 torsion, 576 torsion-free, 532 transformation, 492 unitary, 513 group of homeomorphisms, 101 group topology, 514 Hahn, Hans, 317 Hausdorﬀ-Besicovitch measure, 470 Heine-Borel Theorem, 182 Converse, 183 Generalized, 184 Heisenberg group, 497 hemicompact space, 304 hereditarily separable, 301 Hewitt-Marczewski-Pondiczery Theorem, 289 Higgs boson particle theory, 492 Hilbert cube, 230, 263, 264, 274, 283, 285, 299 Hilbert’s ﬁfth problem, 498 Hilbert, David, 428 Hofmann, Karl Heinrich, 485 homeomorphic, 96 locally, 108 homeomorphism, 96 local, 108 homogeneous, 515 homogeneous coordinates, 491 homomorphism, 494 continuous, 519 INDEX I, 55 identiﬁcation mapping, 343 identiﬁcation spaces, 343 inverse function, 41 image, 42 identity component, 508, 523 inverse element in a group, 493 713 identity matrix, 493 identity of a group, 493 if and only if, 56 image inverse, 42 increasing sequence, 150 indiscrete space, 27 topology, 27 induced topological space, 131 induced topology, 91, 131 induction transﬁnite, 422 inf, 85 inﬁmum, 85 inﬁnite, 394 countably, 394 initial ordinal, 423 initial segment, 420 initial segment topology, 31 injective, 41 injective limit, 533 Int, 83, 164 integer, 55 interior, 83, 164, 526 interior point, 164 Intermediate Value Theorem, 119 intersection of topologies, 46 interval, 104 inverse mapping system, 533 invertible matrix, 493 involution, 388 irrational number, 55 isolated point, 170, 224 isometric, 137, 155 embedding, 155 isometry, 137, 155 isomorphic order, 416 topologically, 496, 519 isomorphic as a C∗-algebra, 388 isomorphism, 494 order, 416 topological, 519 topological group, 519 iterate, 437 jointly continuous, 552 K, 351 k-space, 303 kω-space, 299 kernel, 494 Killing, William, 506 Klein bottle, 351 Klein, Felix, 485 Knuth, Donald, 6, 10 Kronecker, Leopold, 604 1, 159 714 2, 159 ∞, 159 1, 136 2, 136 ∞, 136 La Trobe University, 505 lattice, 379 LCA-group, 523 INDEX Lindenbaum, Adolph, 434 line projective, 490 Sorgenfrey, 84, 179, 197 linear functional, 314 linear group, 513 linear operator, 316 linear order, 268, 415 compactly generated, 582 linear transformation, 525 least element, 85 linearly independent, 271 least upper bound, 379 linearly ordered set, 268–272, 415 Least Upper Bound Axiom, 85 Lipschitz, Rudolf Otto Sigismund , 485 Lebesgue measure, 462, 470, 527 local left uniformity, 551 Lemma homeomorphism, 108 locally Embedding, 238, 274, 277 compact, 189, 207 Urysohn’s, 278 connected, 216 Zorn’s, 179, 269, 269, 271, 272, 318 euclidean, 140 Levine property, 172 Lie group, 496 compact, 496 homeomorphic, 108 locally kω-group, 588 locally compact subgroup, 521, 526 simply connected, 506 locally connected, 252, 301 limit, 628 injective, 533 locally convex space, 210, 211, 265 locally euclidean, 497, 509 projective, 507, 533 locally ﬁnite, 306 limit of a net, 650 limit ordinal, 421 limit point, 73 locally isomorphic, 547 logistic function, 444 Lorentz group, 492, 497 limit point of ﬁlterbase, 642 lower bound, 85, 379 Lindelöf space, 298, 299 Lindelöf ’s Theorem, 243 Lindelöf degree, 312 greatest, 379 lower complete, 312 lower semicontinuous, 172 Lindemann, Ferdinand, 486 Luzin, Nikolai Nikolaevich, 433 INDEX M, 349 m.a.p., 600 Möbius band, 349 Möbius strip, 349 Mandelbrot Benoit, 470 manifold 715 lower semicontinuous, 172 open, 169, 186, 198, 340, 523 quotient, 338 upper semicontinuous, 172 Marczewski, Edward, 285 market stock, 441 Cauchy-Riemann, 140 mathematical induction, 417 connected, 140 CR-, 140 diﬀerentiable, 140, 367, 496 Riemannian, 140 smooth, 140 topological, 140 mathematical proof, 25 matrix identity, 493 invertible, 493 nonsingular, 493 orthogonal, 513 topological with boundary, 140 unitary, 513 MAP, 599 map bijective, 41 conjugate, 466 evaluation, 238 injective, 41 inverse, 41 one-to-one, 41 onto, 41 quadratic, 451 surjective, 41 Maple, 487 mapping closed, 186, 198, 340 continuous, 112 contraction, 161 matrix group, 494 maximal, 269, 269–273 maximally almost periodic, 599 May Robert L., 461 Mazurkiewicz, Stefan, 433 meager, 166 Mean Value Theorem, 163 measure Hausdorﬀ-Besicovitch, 470 Lebesgue, 470, 527 metacompact, 310 metric, 124 bounded, 134 discrete, 125 equivalent, 132 evaluation, 262, 277, 327, 329 euclidean, 125, 125 identiﬁcation, 343 Post Oﬃce, 245 716 space, 124 metric space, 285 bounded, 134 complete, 149 totally bounded, 139 metrizable, 134 -σ, 299 completely, 153 INDEX network, 246 network weight, 246 Neumann, Carl, 485 neutral ﬁxed point, 442 no small subgroups, 509 non-decreasing, 654 non-discrete Hausdorﬀ group topology, 531 metrizable space, 263, 264, 299 nonsingular matrix, 493 minimally almost periodic, 600 norm, 128, 211 monothetic, 580 monotonic sequence, 150 Morris, Sidney A., 505 operator, 316 sup, 370 uniform, 370 Morse normal space, 137, 187, 277, 278, 279, Harold Calvin Marston, 462 283–285, 298–301 Moskowitz-Morris group, 605 normal subgroup, 493, 526 N, 93 N, 26 n-cube, 230 n-sphere, 205 natural numbers Sarkovskii’s ordering, 456 neighboourhood symmetric, 515 neighbourhood, 81 neighbourhood ﬁlter, 621 neighbourhood ﬁlterbase, 641 net, 649 convergent, 650 universal, 653 net associated with ﬁlter, 652 net associated with ﬁlterbase, 652 normed algebra, 377 normed vector space, 128 quotient, 320 nowhere dense, 165 NSS-group, 509 number algebraic, 398 cardinal, 406 irrational, 55 ordinal, 418 prime, 55 transcendental, 398, 486 O(n), 513 ω-narrow, 518, 528 ω, 418 ω1 CK, 423 INDEX object, 110 one-to-one, 41 ordinal Church-Kleene, 423 717 one-to-one correspondence, 393 countable, 423 onto, 41 open ball, 129 covering, 177 initial, 423 limit, 421 predecessor, 421 successor, 421 mapping, 169, 186, 198, 340 ordinal number, 418 sequentially, 146 set, 32 open covering, 177 open mapping, 523 Open Mapping Theorem, 168 for Locally Compact Groups, 535 open subgroup, 520 operator bounded linear, 316 linear, 316 operator norm, 316 or as used in matheamtics, .6 orthogonal matrix, 513 orthogonal group, 494, 513 special, 513 P -based topological space, 313 P, 55, 93 P(S), 401 paracompact, 309 paradox Banach-Tarski, 427 parallelotope, 540 partial derivative, 368 orbit, 437 order linear, 268, 415 partial, 265, 378 strict linear, 268, 415 strict total, 268, 415 order diagram, 267–270 order isomorphic, 416 order isomorphism, 416 order type, 416 ordering partial order, 265, 266, 271, 272, 378 partially ordered set, 265, 266, 268–270, 378, 624 partition of unity, 308 path, 118 path-connected, 118, 209 pathwise connected, 118 peak point, 150 perfect set, 170 perfect space, 170, 224 period doubling bifurcation, 453 Sarkovskii’s, 456 Period Three Theorem, 455 718 periodic, 440 eventually, 440 periodic point attracting, 452 repelling, 452 perspective, 488 Peter, Fritz, 501 phase portrait, 441 Plücker, Julius, 485 plane Sorgenfrey, 301 textbf, 490 Poincarè Jules Henri, 461 point, 73 accumulation, 73 attracting ﬁxed, 442 INDEX Bernstein, 309, 363 complextrignometric, 385 real trignometric, 384 polynomials Bernstein (basis) polynomials, 308 Pontryagin van-Kampen Duality Theorem, 595 Pontryagin, Lev Semyonovich, 502 Pontryagin-van Kampen Duality Theorem for Compact Groups, 575 Pontryagin-van Kampen Duality Theorem for Discrete Groups, 576 population growth, 441, 444 portrait phase, 441 poset, 265, 312 Tukey dominated, 313 attracting periodic, 452 posets boundary, 164 cluster, 73 exterior, 164 ﬁxed, 120, 161 interior, 164 Tukey equivalent, 313 Post Oﬃce metric, 245 power set, 401 pre-annigilator, 320 predecessor ordinal, 421 isolated, 170, 224 prime number, 55 limit, 73 neighbourhood of, 81 neutral ﬁxed, 442 peak, 150 prime period, 440 principal ﬁlter generated by §, 621 Principal Structure Theorem for LCA- Groups, 596 repelling ﬁxed, 442 Principle of the Excluded Middle, 429 repelling periodic, 452 pro-Lie group, 507, 548 pointwise convergent, 365 product, 194, 224, 260, 300 Polish space, 154 polynomial topology, 194 cartesian, 260 INDEX 719 restricted direct, 529 semidirect, 495, 536 space, 194, 260, 261 topology, 200, 260 weak direct, 529 quadratic map, 451 quadrature of the circle, 486 quasicompact, 192 quotient group, 494, 523, 525 quotient mapping, 338 product group, 494 quotient maps not always closed, 524 product of cardinal numbers, 411 quotient normed vector space, 320 product of two ordinals, 419 quotient space, 338 product space, 224, 274, 298 quotient topology, 523 product topology, 63, 224, 260 proﬁnite group, 506 projective dimension, 490 projective geometry, 490 projective group, 495 projective limit, 507, 533 projective line, 490 projective mapping system, 533 R, 31, 51, 265 R2, 60 Rn, 60 R P2, 350 R Pn, 350 projective plane, 490 real, 350 projective space, 490 real, 350 proof by contradiction, 53 if and onl
y if, 56 mathematical, 25 proper subset, 37 property separation, 49 ﬁxed point, 120 topological, 109 protorus, 503 Q, 54, 93, 265 rationally dependent, 538 real projective plane, 350 real projective space, 350 real trignometric polynomial, 384 reduced cone, 347 reduced suspension, 347 reﬁnement, 306, 641 reﬂection, 492 reﬂexive, 97, 578 reﬂexive Banach space, 315, 321, 322 reﬂexive binary relation, 97, 265, 378 reﬂexivity of dual Banach space, 321 regular, 197 completely, 274 space, 95 regular space, 280, 283, 284, 298, 300 720 relation equivalence, 109, 157, 393 relative topology, 91 relatively compact, 188 relativity special, 492 INDEX Sarkovskii’s Theorem, 456 converse, 457 saturated set, 47 scattered space, 245 Schauder, Juliusz Pawel, 434 second axiom of countability, 62, 197, repelling ﬁxed point, 442 241 repelling periodic point, 452 second category, 166 restricted direct product, 529, 576 second countable, 62, 241, 283, 285, 299 retract, 293 Riemannian manifold, 140 right uniformity, 551 rigid motion, 492 Risch, Henery, 487 rotation, 492 Rotkiewicz, Andrzej, 434 Royal Society, 461 Russell, Bertram, 428 Ruziewicz, Stanislaw, 433 SL(n, C), 513 ΣX, 347 σ-compact space, 303 Sn, 350 Sn−1, 344 s-dimensional outer measure seed, 437 semi-open, 108 semicontinuous lower, 172 upper, 172 semidirect product, 536 semidirect product, 495 seminorm, 211 semisimple, 510 sensitively on initial conditions, 464 sensitivity, 464 separable, 83, 154, 197, 236 hereditarily, 301 separable Banach space, 159, 208, 319 Separable Quotient Problem, 323 separable space, 263, 264, 285, 299, 301 separate points, 374 separates points, 238, 262 Hausdorﬀ , 472 separates points and closed sets, 238, S1, 205 S1, 140 Sn, 205 Saks, Stanislaw, 434 262 separation property, 49 sequence Cauchy, 148 Sarkovskii’s ordering, 456 convergent, 141, 190 INDEX decreasing, 150 increasing, 150 monotonic, 150 721 of natural numbers, 26, 93 of positive integers, 26, 93 of rational numbers, 54, 93 sequential space, 146, 303, 341, 342, of real numbers, 31 347 open, 32 sequentially closed, 146 partially ordered, 270 sequentially compact, 190 partially ordered, 265, 266, 268–270, sequentially continuous, 147 sequentially open, 146 set Fσ, 55, 170 Gδ, 55, 170 of continuous real-valued functions, 68 378, 624 perfect, 170 power, 401 saturated, 47 second category, 166 semi-open, 108 totally ordered, 268, 415 absolutely convex, 209 uncountable, 394 absorbent, 209 analytic, 154, 433 bounded, 160 Cantor, 222 clopen, 36 closed, 34 convex, 167, 209 countable, 394 denumerable, 394 derived, 170 directed, 649 ﬁnite, 394 ﬁrst category, 166 inﬁnite, 394 well-ordered, 269, 415 Shrinking Lemma for Normal Spaces, 307 shrinking of a cover, 307 Sierpiński space, 45 Sierpiński curve, 433 carpet, 433 triangle, 433 Sierpiński, Wacław, 432 simple group, 600 Smith P.A., 462 smooth linearly ordered, 268–272, 415 manifold, 140 meager, 166 of integers, 54, 93 smooth function, 366 solenoidal, 598 of irrational numbers, 55, 93 somewhat divisible, 613 722 INDEX Sorgenfrey line, 84, 179, 298, 301 extremally disconnected, 245 Sorgenfrey plane, 301 Souslin space, 154 space T3, 280, 281 T4, 281 σ-compact, 303 σ-metrizable, 299 k-, 303 kω, 588 kω, 299 Tychonoﬀ, 327 Asplund, 324 Baire, 165 ﬁnite, 47 ﬁrst countable, 300 Frechet, 147 Frechet-Urysohnl, 147 Hausdorﬀ, 94, 133, 196, 285, 301, 515 hemicompact, 304 hereditarily separable, 301 homogeneous, 515 identiﬁcation, 343 indiscrete, 27 induced by a metric, 131 Lindelöf, 298, 299 Banach, 158, 265, 525 locally compact, 189, 207 bikompakt, 192 Cantor, 222 locally connected, 216, 252, 301 locally convex, 210, 211, 265 collectionwise Hausdorﬀ, 245 metric, 124, 285 compact, 177, 265, 272–274, 277, metrizable, 134, 263, 264, 299 279, 284, 299, 326, 631, 646 normal, 137, 187, 277–279, 283–285, compactly-generated, 303 298–301 complete metric, 149 completely metrizable, 153 normed vector, 128 perfect, 170, 224 completely regular, 274–276, 300 Polish, 154 connected, 86, 298, 301 product, 194, 224, 260, 261, 274, cosmic, 247 countable-closed, 147 countably compact, 190 disconnected, 87 discrete, 26 door, 47, 94 Dowker, 307 dual, 314 298 projective, 490 quotient, 338 real projective, 350 reﬂexive Banach, 315 regular, 95, 197, 280, 283, 284, 298, 300 scattered, 245 INDEX 723 second countable, 241, 283, 285, 299 special relativity, 492 separable, 83, 154, 197, 236, 263, spread, 312 264, 285, 299, 301 squaring the circle, 486 separable Banach, 159, 208 Steinhaus, Hugo Dyonizy, 427 sequential, 146, 303, 341, 342, 347 stock market, 441 sequentially compact, 190 Sierpiński, 45 Souslin, 154 stronglysσ-metrizable, 299 supercompact, 188 Suslin, 154 T0, 45 T1, 44, 197, 239, 515 T2, 94, 133 T3, 95, 299 T4, 137, 277, 279 , 298, 300, 326 , 274 topological, 25 , 275–277, 279, 299, 327, 329 Stone, Marshall Harvey, 354 Stone-˘Cech Compactiﬁcation, 326, 329– 332, 335 Stone-Weierstrass Theorem, 382, 383 strict linear order, 268, 415 strict total order, 268, 415 strong topology, 315 strongly σ-metrizable, 299 Studia Mathematica, 427 subalgebra, 377 subbasis, 69 subcovering ﬁnite, 177 subgroup, 493 commutator, 526, 533 totally disconnected, 121, 197 locally compact, 521, 526 Tychonoﬀ, 274, 275–277, 279, 298, normal, 493, 526 300, 326, 329 open, 520 uniform, 551 WCG, 322 subgroup topology, 517 sublattice, 379 space of bounded linear operators, 316 subnet, 654 space-ﬁlling curve, 253, 433 space; compact, 326 span, 317 special unitary group, 513 orthogonal group, 513 special linear group, 513 subordinate, 308 subsequence, 150 subset dense, 78 everywhere dense, 78 proper, 37 subspace, 91 724 subspace topology, 91 successor ordinal, 421 T 3 1 2 sum of cardinal numbers, 409 sum of ordinal numbers, 419 T T -space, 274, 275–277, 279, 299, 327, INDEX 329 , 298 -space, 300, 326 sup, 85 Sup norm, 370 supercompact, 188 3 1 2 3 1 2 Tarski, Alfred, 427 tent function, 468 Theorem support of a function, 370 Baire Category, 165 suppose proof by contradiction, 53 supremum, 85 surface, 206 surjective, 41 Suslin space, 154 suspension, 345 reduced, 347 symmetric binary relation, 97 symmetric neighbourhood, 515 symmetry, 492 system dynamical, 463 systems conjugate dynamical, 466 TEX, 6 T0-space, 45 T1-space, 44, 197 T2-space, 94, 133 T3-space, 95 T4-space, 137 T1-space, 239, 515 T3-space, 280, 281, 299 T4-space, 277, 279, 281 Real Stone Weierstrass, 382 Alexander Subbasis, 179 Ascoli, 554 Baire Category, 164 Baire Category for Locally Compact Spaces, 534 Banach Fixed Point, 162 Banach-Alaoglu, 316, 322 Banach-Mazur, 386 Bolzano-Weierstrass, 151 Brouwer Fixed Point, 120 Complex Stone-Weierstrass, 383 Contraction Mapping, 162 Converse of Heine-Borel, 183 Converse of Sarkovskii’s, 457 Dini’s textbf, 370 First Bifurcation, 452 Freyd Adjoint Functor, 326 Fundamental Theorem of Algebra, 217 Gelfand-Naimark Representation, 388 Generalized Heine-Borel, 184, 204 Hahn-Banach, 318, 321 Heine-Borel, 182 Hewitt-Marczewski-Pondiczery, 289 INDEX Lindelöf ’s, 243 Mean Value, 163 Open Mapping, 168 725 topological group, 300, 495, 512 ω-narrow, 518 compactly generated, 522 Open Mapping for Locally Compact connected, 522 Groups, 535 Hausdorﬀ, 527 Pontryagin van-Kampen Duality, 595 isomorphism, 519 Pontryagin-van Kampen Duality for topology, 514 Compact Groups, 575 topological group isomorphism, 519 Pontryagin-van Kampen Duality for topological isomorphism, 519 Discrete Groups, 576 topological property, 109 Principal Structure for LCA-Groups, 596 Sarkovskii’s, 456 The Period Three, 455 Tietze Extension, 294 Toru`nczyk, 256 topological space, 25 P -based, 313 ﬁnite, 47 topological vector space, 208 Frechet, 255 topologically isomorphic, 496, 519 Tychonoﬀ, 203, 259, 265, 273, 631 topologically simple group, 600 Tychonoﬀ ’s, 271, 619 topologically transitive, 466 Urysohn and its Converse, 242 topology, 25 Urysohn Metrization, 284 Urysohn’s, 240, 285 intersection, 46 box, 225 Weierstrass Approximation, 360 coarser, 116, 198 Weierstrass Intermediate Value, 119 coarsest, 261 Well-Ordering, 269, 269, 633 coﬁnite, 39 Well-ordering, 416 compact-open, 552 three space property, 324, 589, 601 countable closed, 45 Tietze Extension Theorem, 294 tightness countable, 147 topological manifold, 140 manifold with boundary, 140 topological ﬁeld, 598 discrete, 26 Euclidean, 621 euclidean, 51 euclidean on Rn, 60 ﬁnal segment, 31 ﬁner, 116, 198 ﬁnite-closed, 39 726 indiscrete, 27 induced, 91 INDEX translation, 492 transpose of a matrix, 494 induced by a metric, 131 triangle initial segment, 31 Sierpiński, 433 product, 63, 194, 200, 224, 260, 260 trisecting an angle, 486 quotient, 523 relative, 91 strong, 315 subgroup, 517 subspace, 91 topological group, 514 Tychonoﬀ, 260 usual, 93 weak, 315 weak, 315 Tukey dominated poset, 313 Tukey equivalent posets, 313 two-sided uniformity, 551 Tychonoﬀ topology, 260 Tychonoﬀ space, 274, 275–277, 279, 298, 300, 326, 327, 329 Tychonoﬀ ’s Theorem, 203, 259, 265, 271, 273, 619, 631 topology of pointwise convergence, 552 torsion group, 576 torsion-free, 532 Toru`nczyk Theorem, 256 torus, 205 totally bounded metric space, 139 U (n), 513 Ulam, Stanislaw, 427 ultraﬁlter, 623, 624, 625 free in C, 660 ultraﬁlter base, 648 ultraﬁlter in C, 660 Ultraﬁlter Lemma, 624 totally disconnected, 121, 197, 524, 526 ultraﬁlterbase, 648 totally disconnected normal subgroup, 526 ultraﬁlterbasis, 648 totally ordered set, 268, 415 ultranet, 653 transcendental number, 398, 486 uncountable set, 394 transﬁnite induction, 422 transformation linear, 525 transformation group, 492 transitive, 463 topologically, 466 uniform norm, 370 uniform space, 551 uniformity, 551 left, 551 right, 551 two-sided, 551 transitive binary relation, 97, 265, 378 uniformly convergent, 365, 371 INDEX union empty, 30 unit ball, 187 unital algebra, 376 unitary matrix, 513 unitary group, 513 special, 513 universal net, 653 727 weak direct product, 529 weak topology, 315 weak-compact, 316 weak*-topology, 315 weakly compactly generated, 322 weather, 441 Weierstra
ss Approximation Theorem, 360 Weierstrass Intermediate Value Theorem, 119 upper bound, 85, 268, 269, 271, 272, Weierstrass, Karl, 354 379 least, 379 weight, 246 network, 246 upper semicontinuous, 172 well-ordered set, 269, 415 Urysohn’s Lemma, 278, 301 well-ordering, 415 Urysohn’s Metrization Theorem, 284 Well-Ordering Theorem, 269, 269, 633 Urysohn’s Theorem, 240, 285 Well-ordering Theorem, 416 Urysohn’s Theorem and its Converse, 242 Weyl, Hermann, 501 usual topology, 93 usual uniformity on R, 551 V -disjoint, 518 Výborný, Rudolf, 354 Wikipedia, 10 Z, 54, 93 Z >, 265 0-dimensional, 122 van der Waerden, Bartel Leendert, 428 Zaremba, Stanislaw, 434 van Kampen, Egbert Rudolf, 502 vanish at inﬁnity, 387 Zentralblatt für Mathematik, 434 Zermelo-Fraenkel axioms, 632 variety of topological groups, 528 zero-dimensional, 122 vector space normed, 128 von Neuman, John, 501 ωX, 663 Wallman compactiﬁcation, 665 Wantzel, Pierre Laurent, 487 WCG space, 322 ZF, 416, 632 ZFC, 416 Zorn’s Lemma, 179, 265, 269, 269, 271, 272, 318
y – A connected graph Ξ. – For each vertex v ∈ V (Ξ), a path-connected space Xv. – For each edge e ∈ E(Ξ), a path-connected space Xe. – For each edge e ∈ E(Ξ) attached to v± ∈ V (Ξ), we have π1-injective maps ∂± e : Xe → Xv± . The realization of X is \|X \| = X = v∈V (Ξ) Xv (∀e ∈ E(Ξ), ∀x ∈ Xe, (x, ±1) ∼ ∂± e∈E(Ξ)(Xe × [−1, 1]) e (x)) . These conditions are not too restrictive. If our vertex or edge space were not path-connected, then we can just treat each path component as a separate vertex/edge. If our maps are not π1 injective, as long as we are careful enough, we can attach 2-cells to kill the relevant loops. Example. A homotopy pushout is a special case of a realization of a graph of spaces. Example. Suppose that the underlying graph looks like this: e v The corresponding gluing diagram looks like Fix a basepoint ∗ ∈ Xe. Pick a path from ∂− loop t that “goes around” Xe × [−1, 1] by ﬁrst starting at ∂− ∗ × [−1, 1], then returning using the path we chose. e (∗) to ∂+ e (∗). This then gives a e (∗), move along Since every loop inside Xe can be pushed down along the “tube” to a loop in Xv, it should not be surprising that the group π1(X) is in fact generated by π1(Xv) and t. In fact, we can explicitly write π1X = π1Xv ∗ t (∂+ e )∗(g) = t(∂− e )∗(g)t−1∀g ∈ π1Xe . This is known as an HNN extension. The way to think about this is as follows — we have a group π1Xv, and we have two subgroups that are isomorphic to each other. Then the HNN extension is the “free-est” way to modify the group so that these two subgroups are conjugate. 24 3 Bass–Serre theory IV Geometric Group Theory How about for a general graph of spaces? If Ξ is a graph of spaces, then its fundamental group has the structure of a graph of groups G. Deﬁnition (Graph of groups). A graph of groups G consists of – A graph Γ – Groups Gv for all v ∈ V (Γ) – Groups Ge for all e ∈ E(Γ) – For each edge e with vertices v±(e), injective group homomorphisms ∂± e : Ge → Gv±(e). In the case of a graph of spaces, it was easy to deﬁne a realization. One way to do so is that we have already see how to do so in the two above simple cases, and we can build the general case up inductively from the simple cases, but that is not so canonical. However, this has a whole lot of choices involved. Instead, we are just going to do is to associate to a graph of groups G a graph of spaces X which “inverts” the natural map from graphs of spaces to graphs of groups, given by taking π1 of everything. This can be done by taking Eilenberg–Maclane spaces. Deﬁnition (Aspherical space). A space X is aspherical if ˜X is contractible. By Whitehead’s theorem and the lifting criterion, this is true iﬀ πn(X) = 0 for all n ≥ 2. Proposition. For all groups G there exists an aspherical space BG = K(G, 1) such that π1(K(G, 1)) ∼= G. Moreover, for any two choices of K(G, 1) and K(H, 1), and for every homomorphism f : G → H, there is a unique map (up to homotopy) ¯f : K(G, 1) → K(H, 1) that induces this homomorphism on π1. In particular, K(G, 1) is well-deﬁned up to homotopy equivalence. Moreover, we can choose K(G, 1) functorially, namely there are choices of K(G, 1) for each G and choices of ¯f such that f1 ◦ f2 = f 1 ◦ ¯f2 and idG = idK(G,1) for all f, G, H. These K(G, 1) are known as Eilenberg–MacLane spaces. When we talked about presentations, we saw that this is true if we don’t have the word “aspherical”. But the aspherical requirement makes the space unique (up to homotopy). Using Eilenberg–MacLane spaces, given any graph of groups G, we can construct a graph of spaces Ξ such that when we apply π1 to all the spaces in X , we recover G. We can now set π1G = π1\|X \|. Note that if Γ is ﬁnite, and all the Gv’s are ﬁnitely-generated, then π1G is also ﬁnitely-generated, which one can see by looking into the construction of K(G, 1). If Γ is ﬁnite, all Gv’s are ﬁnitely-presented and all Ge’s are ﬁnitely-generated, then π1G is ﬁnitely-presented. For more details, read Trees by Serre, or Topological methods in group theory by Scott and Wall. 25 3 Bass–Serre theory IV Geometric Group Theory 3.2 The Bass–Serre tree Given a graph of groups G, we want to analyze π1G = π1X, and we do this by the natural action of G on ˜X by deck transformations. Recall that to understand the free group, we looked at the universal cover of the rose with r petals. Since the rose was a graph, the universal cover is also a graph, and because it is simply connected, it must be a tree, and this gives us a normal form theorem. The key result was that the covering space of a graph is a graph. Lemma. If X is a graph of spaces and ˆX → X is a covering map, then ˆX naturally has the structure of a graph of spaces ˆX , and p respects that structure. Note that the underlying graph of ˆX is not necessarily a covering space of the underlying graph of X . Proof sketch. Consider Xv ⊆ X. v∈V (Ξ) Let   p−1  XV  = ˆWˆv. v∈V (Ξ) ˆv∈V (ˆΞ) This deﬁnes the vertices of ˆΞ, the underlying graph of ˆX . The path components ˆXˆv are going to be the vertex spaces of ˆX . Note that for each ˆv, there exists a unique v ∈ V (Ξ) such that p : ˆXˆv → Xv is a covering map. Likewise, the path components of   p−1 e∈E(Ξ)  Xe × {0}  ˆXˆe of ˆX , which again are covering spaces of the Now let’s deﬁne the edge maps ∂± ˆe for each ˆe ∈ E(ˆΞ) → e ∈ E(Ξ). To do so, form the edge spaces edge space of X. e∈E(Ξ) we consider the diagram ˆXˆe Xe ∼ ∼ ˆXˆe × [−1, 1] Xe × [−1, 1] ˆX p X By the lifting criterion, for the dashed map to exist, there is a necessary and suﬃcient condition on ( ˆXˆe × [−1, 1] → Xe × [−1, 1] → X)∗. But since this condition is homotopy invariant, we can check it on the composition ( ˆXˆe → Xe → X)∗ instead, and we know it must be satisﬁed because a lift exists in this case. The attaching maps ∂± ˆe : ˆXe → ˆX are precisely the restriction to ˆXe×{±1} → ˆX. 26 3 Bass–Serre theory IV Geometric Group Theory Finally, check using covering space theory that the maps ˆXˆe × [−1, 1] → ˆX can injective on the interior of the cylinder, and verify that the appropriate maps are π1-injective. Now let’s apply this to the universal cover ˜X → X. We see that ˜X has a natural action of G = π1X, which preserves the graph of spaces structure. Note that for any graph of spaces X, there are maps such that ρ ◦ ι idΞ. In particular, this implies ρ∗ is surjective (and of course ρ itself is also surjective). In the case of the universal cover ˜X, we see that the underlying graph ˜Ξ = T is connected and simply connected, because π1 ˜Ξ = ρ∗(π1 ˜X) = 1. So it is a tree! The action of G on ˜X descends to an action of G on ˜Ξ. So whenever we have a graph of spaces, or a graph of groups G, we have an action of the fundamental group on a tree. This tree is called the Bass–Serre tree of G. Just like the case of the free group, careful analysis of the Bass–Serre tree leads to theorems relating π1(G) to the vertex groups Gv and edge groups Ge. Example. Let G = F2 = a ∗ b = Z ∗ Z. In this case, we take X to be Xe Xu Xv Here we view this as a graph where the two vertex spaces are circles, and there is a single edge connecting them. The covering space ˜X then looks like This is not the Bass–Serre tree. The Bass–Serre tree is an inﬁnite-valent bipartite tree, which looks roughly like 27 3 Bass–Serre theory IV Geometric Group Theory This point of view gives two important results that relate elements of G to the vertex groups Gvi and the edge maps ∂± ej : Gej → Gv± j . Lemma (Britton’s lemma). For any vertex Ξ, the natural map Gv → G is injective. Unsurprisingly, this really requires that the edge maps are injective. It is an exercise to ﬁnd examples to show that this fails if the boundary maps are not injective. Proof sketch. Observe that the universal cover ˜X can be produce by ﬁrst building universal covers of the vertex space, which are then glued together in a way that doesn’t kill the fundamental groups. More importantly, Bass-Serre trees give us normal form theorems! Pick a base vertex v ∈ V (Ξ). We can then represent elements of G in the form γ = g0e±1 1 g1e±1 2 · · · e±1 n gn where each ei is an edge such that e±1 1 e±1 n forms a closed loop in the 2 underlying graph of Ξ, and each gi is an element of the graph at the vertex connecting e±1 · · · e±1 and e±1 i+1. We say a pinch is a sub-word of the form i e±1∂±1 e (g)e∓1, which can be replaced by ∂∓1 e (g). We say a loop is reduced if it contains no pinches. Theorem (Normal form theorem). Every element can be represented by a reduced loop, and the only reduced loop representing the identity is the trivial loop. This is good enough, since if we can recognize is something is the identity, then we see if the product of one with the inverse of the other is the identity. An exact normal form for all words would be a bit too ambitious. Proof idea. It all boils down to the fact that the Bass–Serre tree is a tree. Connectedness gives the existence, and the simply-connectedness gives the “uniqueness”. 28 4 Hyperbolic groups IV Geometric Group Theory 4 Hyperbolic groups So far, we obtained a lot of information about groups by seeing how they act on trees. Philosophically, the reason for this is that trees are very “negatively curved” spaces. We shall see in this chapter than in general, whenever a group acts on a negatively curved space, we can learn a lot about the group. 4.1 Deﬁnitions and examples We now want to deﬁne a negatively-curved space in great generality. Let X be a geodesic metric space. Given x, y ∈ X, we will write [x, y] for a choice of geodesic between x and y. Deﬁnition (Geodesic triangle). A geodesic triangle ∆ is a choice of three points x, y, z and geodesics [x, y], [y, z], [z, x]. Geodesic triangles look like this: Note that in general, the geodesics may intersect. Deﬁnition (δ-slim triangle). We say ∆ is δ-slim if every side of ∆ is contained in the union of the δ-neighbourhoods of the other two sides. Deﬁnition (Hyperbolic space). A metric space is (Gromov) hyperbolic if there exists δ ≥ 0 such that every
geodesic triangle in X is δ-slim. In this case, we say it is δ-hyperbolic. Example. R2 is not Gromov hyperbolic. Example. If X is a tree, then X is 0-hyperbolic! Indeed, each triangle looks like We call this a tripod . Unfortunately, none of these examples really justify why we call these things hyperbolic. Let’s look at the actual motivating example. Example. Let X = H2, the hyperbolic plane. Let ∆ ⊆ H2 be a triangle. Then ∆ is δ-slim, where δ is the maximum radius of an inscribed semi-circle D in ∆ with the center on one of the edges. But we know that the radius of D is bounded by some increasing function of the area of D, and the area of D is bounded above by the area of ∆. On the other hand, by hyperbolic geometry, we know the area of any triangle is bounded by π. So H2 is δ-hyperbolic for some δ. 29 4 Hyperbolic groups IV Geometric Group Theory If we worked a bit harder, then we can ﬁgure out the best value of δ. However, for the arguments we are going to do, we don’t really care about what δ is. Example. Let X be any bounded metric space, e.g. S2. Then X is Gromov hyperbolic, since we can just take δ to be the diameter of the metric space. This is rather silly, but it makes sense if we take the “coarse point of view”, and we have to ignore bounded things. What we would like to do is to say is that a group Γ if for every ﬁnite generating set S, the Cayley graph CayS(Γ) equipped with the word metric is δ-hyperbolic for some δ. However, this is not very helpful, since we have to check it for all ﬁnite generating sets. So we want to say that being hyperbolic is quasi-isometry invariant, in some sense. This is slightly diﬃcult, because we lose control of how the geodesic behaves if we only look at things up to isometry. To do so, we have to talk about quasi-geodesics. 4.2 Quasi-geodesics and hyperbolicity Deﬁnition (Quasi-geodesic). A (λ, ε)-quasi-geodesic for λ ≥ 1, ε ≥ 0 is a (λ, ε)-quasi-isometric embedding I → X, where I ⊆ R is a closed interval. Note that our deﬁnition allows for I to be unbounded, i.e. I may be of the forms [a, b], [0, ∞) or R respectively. We call these quasi-geodesic intervals, quasi-geodesic rays and quasi-geodesic lines respectively. Example. The map [0, ∞) → R2 given in polar coordinates by is a quasigeodesic ray. t → (t, log(1 + t)) This should be alarming. The quasi-geodesics in the Euclidean plane can look unlike any genuine geodesic. However, it turns out things are well-behaved in hyperbolic spaces. Theorem (Morse lemma). For all δ ≥ 0, λ ≥ 1 there is R(δ, λ, ε) such that the following holds: If X is a δ-hyperbolic metric space, and c : [a, b] → X is a (λ, ε)-quasigeodesic from p to q, and [p, q] is a choice of geodesic from p to q, then dHaus([p, q], im(c)) ≤ R(δ, λ, ε), where dHaus(A, B) = inf{ε > 0 \| A ⊆ Nε(B) and B ⊆ Nε(A)} is the Hausdorﬀ distance. This has nothing to do with the Morse lemma in diﬀerential topology. Corollary. There is an M (δ, λ, ε) such that a geodesic metric space X is δhyperbolic iﬀ any (λ, ε)-quasigeodesic triangle is M -slim. 30 4 Hyperbolic groups IV Geometric Group Theory Corollary. Suppose X, X are geodesic metric spaces, and f : X → X is a quasi-isometric embedding. If X is hyperbolic, then so is X. In particular, hyperbolicity is a quasi-isometrically invariant property, when restricted to geodesic metric spaces. Thus, we can make the following deﬁnition: Deﬁnition (Hyperbolic group). A group Γ is hyperbolic if it acts properly discontinuously and cocompactly by isometries on a proper, geodesic hyperbolic metric space. Equivalently, if it is ﬁnitely-generated and with a hyperbolic Cayley graph. Let’s prove the Morse lemma. To do so, we need the following deﬁnition: Deﬁnition (Length of path). Let c : [a, b] → X be a continuous path. Let a = t0 < t1 < · · · < tn = b be a dissection D of [a, b]. Deﬁne (c) = sup D n i=1 d(c(ti−1), c(ti)). In general, this number is extremely hard to compute, and may even be inﬁnite. Deﬁnition (Rectiﬁable path). We say a path c is rectiﬁable if (c) < ∞. Example. Piecewise geodesic paths are rectiﬁable. Lemma. Let X be a geodesic space. For any (λ, ε)-quasigeodesic c : [a, b] → X, there exists a continuous, rectiﬁable (λ, ε)-quasigeodesic c : [a, b] → X with ε = 2(λ + ε) such that (i) c(a) = c(a), c(b) = c(b). (ii) For all a ≤ t < t ≤ b, we have (c\|[t,t]) ≤ k1d(c(t), c(t)) + k2 where k1 = λ(λ + ε) and k2 = (λε + 3)(λ + 3). (iii) dHaus(im c, im c) ≤ λ + ε. Proof sketch. Let Σ = {a, b} ∪ ((a, b) ∩ Z). For t ∈ Σ, we let c(t) = c(t), and deﬁne c to be geodesic between the points of Σ. Then claims (i) and (iii) are clear, and to prove quasigeodesicity and (ii), let σ : [a, b] → Σ be a choice of closest point in Σ, and then estimate d(c(t), c(t)) in terms of d(c(σ(t)), c(σ(t))). Lemma. let X be δ-hyperbolic, and c : [a, b] → X a continuous, rectiﬁable path in X joining p to q. For [p, q] a geodesic, for any x ∈ [p, q], we have d(x, im c) ≤ δ\| log2 (c)\| + 1. Proof. We may assume c : [0, 1] → X is parametrized proportional to arc length. Suppose (c) 2N < 1 ≤ (c) 2N −1 . Let x0 = x. Pick a geodesic triangle between p, q, c( 1 2 ). By δ-hyperbolicity, there exists a point x1 lying on the other two edges such that d(x0, x1) ≤ δ. We wlog assume x1 ∈ [p, c( 1 2 ]. 2 )]. We can repeat the argument with c\|[0, 1 31 4 Hyperbolic groups IV Geometric Group Theory c( 1 2 ) x1 p x0 q Formally, we proceed by induction on N . If N = 0 so that (c) < 1, then we are done by taking desired point on im c to be p (or q). Otherwise, there is some x1 ∈ [p, c( 1 2 )] such that d(x0, x1) ≤ δ. Then (c\|[0, 1 2 ]) 2N −1 < 1 ≤ (c\|[0, 1 2N −2 2 ]) . So by the induction hypothesis, d(x1, im c) ≤ δ\| log2 (c\|[0, 1 2 ]\| + 1 log2 (c) = δ + 1 1 2 = δ(\|log2 (c)\| − 1) + 1. Note that we used the fact that (c) > 1, so that log2 (c) > 0. Then we are done since d(x, im c) ≤ d(x, x1) + d(x1, im c). Proof of Morse lemma. By the ﬁrst lemma, we may assume that c is continuous and rectiﬁable, and satisﬁes the properties as in the lemma. Let p, q be the end points of c, and [p, q] a geodesic. First we show that [p, q] is contained in a bounded neighbourhood of im c. Let D = sup x∈[p,q] d(x, im c). By compactness of the interval, let x0 ∈ [p, q] where the supremum is attained. Then by assumption, im c lies outside ˚B(x0, D). Choose y, z ∈ [p, q] be such that d(x0, y) = d(x0, z) = 2D and y, x0, z appear on the geodesic in this order (take y = p, or z = q if that is not possible). Let y = c(s) ∈ im c be such that d(y, y) ≤ D, and similarly let z = c(t) ∈ im c be such that d(z, z) ≤ D. y p ≤ D y D x0 32 2D c z q ≤ D z 2D 4 Hyperbolic groups IV Geometric Group Theory Let γ = [y, y] · c\|[s,t] · [z, z]. Then (γ) = d(y, y) + d(z, z) + (c\|[s,t]) ≤ D + D + k1d(y, z) + k2, by assumption. Also, we know that d(y, z) ≤ 6D. So we have But we know that (γ) ≤ 6k1D + 2D + k2. d(x0, im γ) = D. So the second lemma tells us D ≤ δ\| log2(6k1D + 2D + k2)\| + 1. The left hand side is linear in D, and the right hand side is a logarithm in D. So it must be the case that D is bounded. Hence [p, q] ⊆ ND0(im c), where D0 is some constant. It remains to ﬁnd a bound M such that im c ⊆ NM ([p, q]). Let [a, b] be a maximal subinterval of [a, b] such that c[a, b] lies entirely outside ˚ND0 ([p, q]). Since ¯ND(c[a, a]) and ¯ND(c[b, b]) are both closed, and they collectively cover the connected set [p, q], there exists w ∈ [p, q] ∩ ¯ND0 (c[a, a]) ∩ ¯ND0(c[b, b]). Therefore there exists t ∈ [a, a] and t ∈ [b, b] such that d(w, c(t)) ≤ D0 and d(w, c(t)) ≤ D0. In particular, d(c(t), c(t)) ≤ 2D0. By the ﬁrst lemma, we know (c\|[t,t]) ≤ 2k1D0 + k2. So we know that for s ∈ [a, b], we have d(c(s), [p, q]) ≤ d(c(s), w) ≤ d(c(s), c(t)) + d(c(t), w) ≤ (c\|[t,t]) + D0 ≤ D0 + 2k1D0 + k2. With the Morse lemma proven, we can sensibly talk about hyperbolic groups. We have already seen many examples of hyperbolic spaces, such as trees and the hyperbolic plane. Example. Any ﬁnite group is hyperbolic since they have bounded Cayley graphs. Example. Finitely-generated free groups are hyperbolic, since their Cayley graphs are trees. Example. Surface groups Σg for g ≥ 2 acts on the hyperbolic plane properly discontinuously and cocompactly by isometries (since the universal cover of the genus g surface is the hyperbolic plane by tessellating the hyperbolic plane with 4g-gons, cf. IB Geometry). The following notion of “virtually” would be helpful 33 4 Hyperbolic groups IV Geometric Group Theory Deﬁnition (virtually-P). Let P be a property of groups. Then we say G is virtually-P if there is a ﬁnite index subgroup G0 ≤ G such that G0 is P. Example. Finite groups are virtually trivial! Note that G0 ≤ G is ﬁnite-index, then G0 acts on CayS(G) in a way that satisﬁes the properties of the Schwarz–Milnor lemma (the only property we may worry about is cocompactness, which is what we need ﬁnite-index for) and hence G0 qi G. In particular, Example. A virtually hyperbolic group is hyperbolic. For example, virtually free groups such that (Z/2) ∗ (Z/3) = PSL2Z are hyperbolic. These are some nice classes of examples, but they will be dwarfed by our next class of examples. A “random group” is hyperbolic. More precisely, ﬁx m ≥ 2 and n ≥ 1. We temporarily ﬁx ≥ 0. Consider groups of the form Γ = a1, . . . , am \| r1, . . . , rn. such that ri are all cyclicly reduced words of length (i.e. reduced words such that the ﬁrst letter is not the inverse of the last). Put the uniform probability distribution on the set of all such groups. This deﬁnes a group-valued random variable Γ. For a property P , we say that “a random group is P ” if lim →∞ P(Γ is hyperbolic) = 1 for all n, m. Theorem (Gromov). A random group is inﬁnite and hyperbolic. There are, of course, other ways to deﬁne a “random group”. As long as we control the number of relations well so that we don’t end up with ﬁnite groups all the time, the “random group” should still be hyperbolic. Recall that in diﬀerential geometry, geodesics are deﬁned locally. However, we deﬁned our geodesics to be emb
edded interval, which is necessarily a global notion. We want an analogous local version. However, if we want to work up to quasi-isomorphism, then we cannot go completely local, because locally, you are allowed to be anything. Deﬁnition (k-local geodesic). Let X be a geodesic metric space, and k > 0. A path c : [a, b] → X is a k-local geodesic if d(c(t), c(t)) = \|t − t\| for all t, t ∈ [a, b] with \|t − t\| ≤ k. We know very well that on a general Riemannian manifold, local geodesics need not be actual geodesics. In fact, they can be far from being an actual geodesic, since, for example, we can wrap around the sphere many times. Again, things are much better in the hyperbolic world. Theorem. Let X be δ-hyperbolic and c : [a, b] → X be a k-local geodesic where k > 8δ. Then c is a (λ, ε)-quasigeodesic for some λ = λ(δ, k) and ε = ε(δ, k). 34 4 Hyperbolic groups IV Geometric Group Theory First, we prove Lemma. Let X be δ-hyperbolic and k > 8δ. geodesic, then im c is contained in the 2δ-neighbourhood of [c(a), c(b)]. If c : [a, b] → X is a k-local Observe that by iterating the deﬁnition of hyperbolicity, on a δ-hyperbolic space, any point on an n-gon is at most (n − 2)δ away from a point on another side. Proof. Let x = c(t) maximize d(x, [c(a), c(b)]). Let If t − k 2 < a, we set y = c(a) instead, and similarly for z. Let y ∈ [c(a), c(b)] minimize d(y, y), and likewise let z ∈ [c(a), c(b)] minimize d(z, z). y x w y z z Fix geodesics [y, y] and [z, z]. Then we have a geodesic rectangle with vertices y, y, z, z. By δ-hyperbolicity, there exists w on the rectangle not on im c, such that d(x, w) = 2δ. If w ∈ [y, z], then we win. Otherwise, we may wlog assume w ∈ [y, y]. Note that in the case y = c(a), we must have y = y, and so this would imply w = y ∈ [c(a), c(b)]. So we are only worried about the case y = c t − k . So d(y, x) = k 2 > 4δ. But then by the triangle inequality, we must have d(y, w) > 2δ = d(x, w). 2 However, d(x, y) ≤ d(x, w) + d(w, y) < d(y, w) + d(w, y) = d(y, y). So it follows that d(x, [c(a), c(b)]) < d(y, y) = d(y, [c(a), c(b)]). This contradicts our choice of x. Proof of theorem. Let c : [a, b] → X be a k-local geodesic, and t ≤ t ∈ [a, b]. Choose t0 = t < t1 < · · · < tn < t such that ti = ti−1 + k for all i and t − tn < k. Then by deﬁnition, we have d(c(ti−1), c(ti)) = k, d(c(tn), c(t)) = \|tn − t\|. 35 4 Hyperbolic groups IV Geometric Group Theory for all i. So by the triangle inequality, we have d(c(t), c(t)) ≤ n i=1 d(c(ti−1), c(ti)) + d(tn, t) = \|t − t\|. We now have to establish a coarse lower bound on d(c(t), c(t)). We may wlog assume t = a and t = b. We need to show that d(c(a), c(b)) ≥ 1 λ \|b − a\| − ε. We divide c up into regular subintervals [xi, xi+1], and choose x goal is then to prove that the x i appear in order along [c(a), c(b)]. i close to xi. The Let k = k 2 + 2δ > 6δ. Let b − a = M k + η for 0 ≤ η < k and M ∈ N. Put xi = c(ik) for i = 1, . . . , M , and let x i be a closest point on [c(a), c(b)] to xi. By the lemma, we know d(xi, x Claim. x m appear in the correct order along [c(a), c(b)]. 1, . . . , x i) ≤ 2δ. Let’s ﬁnish the proof assuming the claim. If this holds, then note that d(x i, x i+1) ≥ k − 4δ > 2δ because we know d(xi, xi+1) = 6δ, and also d(xm, c(b)) ≥ η − 2δ. Therefore d(c(a), c(b)) = M i=1 d(xi, xi−1) + d(xm, c(b)) ≥ 2δM + η − 2δ ≥ 2δ(M − 1). On the other hand, we have M = \|b − a\| − η k ≥ \|b − a\| k − 1. Thus, we ﬁnd that 2δ k \|b − a\| − 4δ. To prove the claim, let xi = c(ti) for all i. We let d(c(a), c(b)) ≥ Deﬁne y = c(ti−1 + 2δ) z = c(ti+1 − 2δ). ∆− = ∆(xi−1, x ∆+ = ∆(xi+1, x i−1, y) i+1, z). Both ∆− and ∆+ are disjoint from B(xi, 3δ). Indeed, if w ∈ ∆− with d(xi, w) ≤ 3δ, then by δ-slimness of ∆−, we know d(w, xi−1) ≤ 3δ, and so d(xi, xi−1) ≤ 6δ, which is not possible. 36 4 Hyperbolic groups IV Geometric Group Theory Therefore, since the rectangle y, z, x 2δ away from the sides yx with d(xi, x i ) ≤ 2δ. Now consider ∆ = ∆(xi, x i−1 and zx i+1, x i+1. So there must be some x i−1 is 2δ-slim, and xi is more than i+1] i ∈ [x i−1, x i, x i ). We know xix i are both of length ≤ 2δ. Note that every point in this triangle is within 3δ of xi by δ-slimness. So ∆ ⊆ B(xi, 3δ), and this implies ∆ is disjoint from B(xi−1, 3δ) and B(xi+1, 3δ) as before. But x i−1 ∈ B(xi−1, 3δ) and x i+1 ∈ B(xi+1, 3δ). Moreover, ∆ contains the i . Therefore, it must be the case that i and xix i and x segment of [c(a), c(b)] joining x i ∈ [x x i−1, x i+1]. 4.3 Dehn functions of hyperbolic groups We now use our new understanding of quasi-geodesics in hyperbolic spaces to try to understand the word problem in hyperbolic groups. Note that by the Schwarz–Milnor lemma, hyperbolic groups are ﬁnitely-generated and their Cayley graphs are hyperbolic. Corollary. Let X be δ-hyperbolic. Then there exists a constant C = C(δ) such that any non-constant loop in X is not C-locally geodesic. Proof. Take k = 8δ + 1, and let C = max{λε, k} where λ, ε are as in the theorem. Let γ : [a, b] → X be a closed loop. If γ were C-locally geodesic, then it would be (λ, ε)-quasigeodesic. So So 0 = d(γ(a), γ(b)) ≥ \|b − a\| λ − ε. \|b − a\| ≤ λε < C. But γ is a C-local geodesic. This implies γ is a constant loop. Deﬁnition (Dehn presentation). A ﬁnite presentation S \| R for a group Γ is called Dehn if for every null-homotopic reduced word w ∈ S∗, there is (a cyclic conjugate of) a relator r ∈ R such that r = u−1v with S(u) < S(v), and w = w1vw2 (without cancellation). The point about this is that if we have a null-homotopic word, then there is some part in the word that can be replaced with a shorter word using a single relator. If a presentation is Dehn, then the naive way of solving the word problem just works. In fact, Lemma. If Γ has a Dehn presentation, then δΓ is linear. Proof. Exercise. Theorem. Every hyperbolic group Γ is ﬁnitely-presented and admits a Dehn presentation. In particular, the Dehn function is linear, and the word problem is solvable. 37 4 Hyperbolic groups IV Geometric Group Theory So while an arbitrary group can be very diﬃcult, the generic group is easy. Proof. Let S be a ﬁnite generating set for Γ, and δ a constant of hyperbolicity for CayS(Γ). Let C = C(δ) be such that every non-trivial loop is not C-locally geodesic. Take {ui} to be the set of all words in F (S) representing geodesics [1, ui] in CayS(Γ) with \|ui\| < C. Let {vj} ⊆ F (S) be the set of all non-geodesic words of length ≤ C in CayS(Γ). Let R = {u−1 i vj ∈ F (S) : ui =Γ vj}. We now just observe that this gives the desired Dehn presentation! Indeed, any non-trivial loop must contain one of the vj’s, since CayS(Γ) is not C-locally geodesic, and re can replace it with ui! This argument was developed by Dehn to prove results about the fundamental group of surface groups in 1912. In the 1980’s, Gromov noticed that Dehn’s argument works for an arbitrary hyperbolic group! One can keep on proving new things about hyperbolic groups if we wished to, but there are better uses of our time. So for the remaining of the chapter, we shall just write down random facts about hyperbolic groups without proof. So hyperbolic groups have linear Dehn functions. In fact, Theorem (Gromov, Bowditch, etc). If Γ is a ﬁnitely-presented group and δΓ ň n2, then Γ is hyperbolic. Thus, there is a “gap” in the isoperimetric spectrum. We can collect our results as Theorem. If Γ is ﬁnitely-generated, then the following are equivalent: (i) Γ is hyperbolic. (ii) Γ has a Dehn presentation. (iii) Γ satisﬁes a linear isoperimetric inequality. (iv) Γ has a subquadratic isoperimetric inequality. In general, we can ask the question — for which α ∈ R is nα a Dehn function of a ﬁnitely-presented group? As we saw, α cannot lie in (1, 2), and it is a theorem that the set of such α is dense in [2, ∞). In fact, it includes all rationals in the interval. Subgroup structure When considering subgroups of a hyperbolic group Γ, it is natural to consider “geometrically nice” subgroups, i.e. ﬁnitely-generated subgroups H ⊆ Γ such that the inclusion is a quasi-isometric embedding. Such subgroups are called quasi-convex , and they are always hyperbolic. What sort of such subgroups can we ﬁnd? There are zillions of free quasi- convex subgroups! Lemma (Ping-pong lemma). Let Γ be hyperbolic and torsion-free (for convenience of statement). If γ1, γ2 ∈ Γ do not commute, then for large enough n, the subgroup γn 2 ∼= F2 and is quasi-convex. 1 , γn 38 4 Hyperbolic groups IV Geometric Group Theory How about non-free subgroups? Can we ﬁnd surface groups? Of course, we cannot always guarantee the existence of such surface groups, since all subgroups of free groups are free. Question. Let Γ be hyperbolic and torsion-free, and not itself free. Must Γ contain a quasi-convex subgroup isomorphic to π1Σ for some closed hyperbolic surface Σ? We have no idea if it is true or false. Another open problem we can ask is the following: Question. If Γ is hyperbolic and not the trivial group, must Γ have a proper subgroup of ﬁnite index? Proposition. Let Γ be hyperbolic, and γ ∈ Γ. Then C(γ) is quasiconvex. In particular, it is hyperbolic. Corollary. Γ does not contain a copy of Z2. The boundary Recall that if Σ is a compact surface of genus g ≥ 2, then π1Σ qi H2. If we try to draw the hyperbolic plane in the disc model, then we would probably draw a circle and ﬁll it in. One might think the drawing of the circle is just an artifact of the choice of the model, but it’s not! It’s genuinely there. Deﬁnition (Geodesic ray). Let X be a δ-hyperbolic geodesic metric space. A geodesic ray is an isometric embedding r : [0, ∞) → X. We say r1 ∼ r2 if there exists M such that d(r1(t), r2(t)) ≤ M for all t. In the disc model of H2, this is the scenario where two geodesic rays get very close together as t → ∞. For example, in the upper half plane model of H2, all vertical lines are equivalent in this sense. We deﬁne ∂∞X = {geodesic rays}/ ∼. This can be topologized in a sensible way, and in this case X ∪ ∂∞X is compact. By the Morse lemma, for
hyperbolic spaces, this is quasi-isometry invariant. Example. If Γ = π1Σ, with Σ closed hyperbolic surface, then ∂∞Γ = S1 and the union X ∪ ∂∞X gives us the closed unit disc. Theorem (Casson–Jungreis, Gabai). If Γ is hyperbolic and ∂∞Γ ∼= S1, then Γ is virtually π1Σ for some closed hyperbolic Σ. Example. If Γ is free, then ∂∞Γ is the Cantor set. Conjecture (Cannon). If Γ is hyperbolic and ∂∞Γ ∼= S2, then Γ is virtually π1M for M a closed hyperbolic 3-manifold. 39 5 CAT(0) spaces and groups IV Geometric Group Theory 5 CAT(0) spaces and groups From now on, instead of thinking of geodesics as being isometric embeddings, we reparametrize them linearly so that the domain is always [0, 1]. 5.1 Some basic motivations Given a discrete group Γ, there are two basic problems you might want to solve. Question. Can we solve the word problem in Γ? Question. Can we compute the (co)homology of Γ? Deﬁnition (Group (co)homology). The (co)homology of a group Γ is the (co)homology of K(Γ, 1). We can deﬁne this in terms of the group itself, but would require knowing some extra homological algebra. A very closely related question is Question. Can we ﬁnd an explicit X such that Γ = π1X and ˜X is contractible? We know that these problems are not solvable in general: Theorem (Novikov–Boone theorem). There exists a ﬁnitely-presented group with an unsolvable word problem. Theorem (Gordon). There exists a sequence of ﬁnitely generated groups Γn such that H2(Γn) is not computable. As before, we might expect that we can solve these problems if our groups come with some nice geometry. In the previous chapter, we talked about hyperbolic groups, which are negatively curved. In this section, we shall work with slightly more general spaces, namely those that are non-positively curved. Let M be a compact manifold of non-positive sectional curvature. It is a classical fact that such a manifold satisﬁes a quadratic isoperimetric inequality. This is not too surprising, since the “worst case” we can get is a space with constant zero curvature, which implies ˜M ∼= Rn. If we know this, then by the Filling theorem, we know the Dehn function of the fundamental group is at worst quadratic, and in particular it is computable. This solves the ﬁrst question. What about the second question? Theorem (Cartan–Hadamard theorem). Let M be a non-positively curved compact manifold. Then ˜M is diﬀeomorphic to Rn. In particular, it is contractible. Thus, M = K(π1M, 1). For example, this applies to the torus, which is not hyperbolic. So non-positively curved manifolds are good. However, there aren’t enough of them. Why? In general, the homology of a group can be very complicated, and in particular can be inﬁnite dimensional. However, manifolds always have ﬁnite-dimensional homology groups. Moreover, they satisfy Poincar´e duality. Theorem (Poincar´e duality). Let M be an orientable compact n-manifold. Then Hk(M ; R) ∼= Hn−k(M ; R). 40 5 CAT(0) spaces and groups IV Geometric Group Theory This is a very big constraint, and comes very close to characterizing manifolds. In general, it is diﬃcult to write down a group whose homology satisﬁes Poincar´e duality, unless we started oﬀ with a manifold whose universal cover is contractible, and then took its fundamental group. Thus, we cannot hope to realize lots of groups as the π1 of a non-positively curved manifold. The idea of CAT(0) spaces is to mimic the properties of non-positively curved manifolds in a much more general setting. 5.2 CAT(κ) spaces Let κ = −1, 0 or 1, and let Mκ be the unique, simply connected, complete 2-dimensional Riemannian manifold of curvature κ. Thus, M1 = S2, M0 = R2, M−1 = H2. We can also talk about Mκ for other κ, but we can just obtain those by scaling M±1. Instead of working with Riemannian manifolds, we shall just think of these as complete geodesic metric spaces. We shall now try to write down a “CAT(κ)” condition, that says the curvature is bounded by κ in some space. Deﬁnition (Triangle). A triangle with vertices {p, q, r} ⊆ X is a choice ∆(p, q, r) = [p, q] ∪ [q, r] ∪ [r, p]. If we want to talk about triangles on a sphere, then we have to be a bit more careful since the sides cannot be too long. Let Dκ = diam Mκ, i.e. Dκ = ∞ for κ = 0, −1 and Dκ = π for κ = +1. Suppose ∆ = ∆(x1, x2, x3) is a triangle of perimeter ≤ 2Dκ in some complete geodesic metric space (X, d). Then there is, up to isometry, a unique comparison triangle ¯∆ = ∆(¯x1, ¯x2, ¯x3) ⊆ Mκ such that dMk (¯xi, ¯xj) = d(xi, xj). This is just the fact we know from high school that a triangle is determined by the lengths of its side. The natural map ¯∆ → ∆ is called the comparison triangle. Similarly, given a point p ∈ [xi, xj], there is a comparison point ¯p ∈ [¯xi, ¯yj]. Note that p might be on multiple edges, so it could have multiple comparison points. However, the comparison point is well-deﬁned as long as we specify the edge as well. Deﬁnition (CAT(κ) space). We say a space (X, d) is CAT(κ) if for any geodesic triangle ∆ ⊆ X of diameter ≤ 2Dκ, any p, q ∈ ∆ and any comparison points ¯p, ¯q ∈ ¯∆, d(p, q) ≤ dMκ (¯p, ¯q). If X is locally CAT(κ), then K is said to be of curvature at most κ. In particular, a locally CAT(0) space is called a non-positively curved space. We are mostly interested in CAT(0) spaces. At some point, we will talk about CAT(1) spaces. 41 5 CAT(0) spaces and groups IV Geometric Group Theory Example. The following are CAT(0): (i) Any Hilbert space. (ii) Any simply-connected manifold of non-positive sectional curvature. (iii) Symmetric spaces. (iv) Any tree. (v) If X, Y are CAT(0), then X × Y with the 2 metric is CAT(0). (vi) In particular, a product of trees is CAT(0). Lemma (Convexity of the metric). Let X be a CAT(0) space, and γ, δ : [0, 1] → X be geodesics (reparameterized). Then for all t ∈ [0, 1], we have d(γ(t), δ(t)) ≤ (1 − t)d(γ(0), δ(0)) + td(γ(1), δ(1)). Note that we have strict equality if this is in Euclidean space. So it makes sense that in CAT(0) spaces, we have an inequality. Proof. Consider the rectangle δ(0) γ(0) δ γ α δ(1) γ(1) Let α : [0, 1] → X be a geodesic from γ(0) to δ(1). Applying the CAT(0) estimate to ∆(γ(0), γ(1), δ(1)), we get d(γ(t), α(t)) ≤ d(γ(t), α(t)) = td(γ(1), α(1)) = td(γ(1), α(1)) = td(γ(1), δ(1)), using what we know in plane Euclidean geometry. The same argument shows that d(δ(t), α(t)) ≤ (1 − t)d(δ(0), γ(0)). So we know that d(γ(t), δ(t)) ≤ d(γ(t), α(t)) + d(α(t), δ(t)) ≤ (1 − t)d(γ(0), δ(0)) + td(γ(1), δ(1)). Lemma. If X is CAT(0), then X is uniquely geodesic, i.e. each pair of points is joined by a unique geodesic. Proof. Suppose x0, x1 ∈ X and γ(0) = δ(0) = x0 and γ(1) = δ(1) = x1. Then by the convexity of the metric, we have d(γ(t), δ(t)) ≤ 0. So γ(t) = δ(t) for all t. This is not surprising, because this is true in the Euclidean plane and the hyperbolic plane, but not in the sphere. Of course, one can also prove this result directly. 42 5 CAT(0) spaces and groups IV Geometric Group Theory Lemma. Let X be a proper, uniquely geodesic metric space. Then geodesics in X vary continuously with their end points in the compact-open topology (which is the same as the uniform convergence topology). This is actually true without the word “proper”, but the proof is harder. Proof. This is an easy application of the Arzel´a–Ascoli theorem. Proposition. Any proper CAT(0) space X is contractible. Proof. Pick a point x0 ∈ X. Then the map X → Maps([0, 1], X) sending x to the unique geodesic from x0 to x is continuous. The adjoint map X × [0, 1] → X is then a homotopy from the constant map at x0 to the identity map. Deﬁnition (CAT(0) group). A group is CAT(0) if it acts properly discontinuously and cocompactly by isometries on a proper CAT(0) space. Usually, for us, the action will also be free. This is the case, for example, when a fundamental group acts on the covering space. Note that a space being CAT(0) is a very ﬁne-grained property, so this is not the same as saying the Cayley graph of the group is CAT(0). Example. Zn for any n is CAT(0), since it acts on Rn. Example. More generally, π1M for any closed manifold M of non-positive curvature is CAT(0). Example. Uniform lattices in semi-simple Lie groups. Examples include SLnOK for certain number ﬁelds K. Example. Any free group, or direct product of free groups is CAT(0). We remark, but will not prove Proposition. Any CAT(0) group Γ satisﬁes a quadratic isoperimetric inequality, that is δΓ n or ∼ n2. Note that if Γ is in fact CAT(-1), then Γ is hyperbolic, which is not terribly diﬃcult to see, since H2 is hyperbolic. But if a group is hyperbolic, is it necessarily CAT(-1)? Or even just CAT(0)? This is an open question. The diﬃculty in answering this question is that hyperbolicity is a “coarse condition”, but being CAT(0) is a ﬁne condition. For example, Cayley graphs are not CAT(0) spaces unless they are trees. 5.3 Length metrics In diﬀerential geometry, if we have a covering space ˜X → X, and we have a Riemannian metric on X, then we can lift this to a Riemannian metric to ˜X. This is possible since the Riemannian metric is a purely local notion, and hence we can lift it locally. Can we do the same for metric spaces? Recall that if γ : [a, b] → X is a path, then (γ) = sup a=t0<t1<···<tn=b n i=1 d(γ(ti−1), γ(ti)). We say γ is rectiﬁable if (γ) < ∞. 43 5 CAT(0) spaces and groups IV Geometric Group Theory Deﬁnition (Length space). A metric space X is called a length space if for all x, y ∈ X, we have d(x, y) = inf γ:x→y (γ). Given any metric space (X, d), we can construct a length pseudometric ˆd : X × X → [0, ∞] given by ˆd(x, y) = inf γ:x→y (γ). Now given a covering space p : ˜X → X and (X, d) a metric space, we can deﬁne a length pseudometric on ˜X by, for any path ˜γ : [a, b] → ˜X, (˜γ) = (p ◦ ˜γ). This induces an induced pseudometric on ˜X. Exercise. If X is a length space, then so is ˜X. Note that if A, B are length spaces, and X = A ∪ B (such that the metrics agree on the intersection), then X has a natural induced length
metric. Recall we previously stated the Hopf–Rinow theorem in the context of diﬀerential geometry. In fact, it is actually just a statement about length spaces. Theorem (Hopf–Rinow theorem). If a length space X is complete and locally compact, then X is proper and geodesic. This is another application of the Arzel´a–Ascoli theorem. 5.4 Alexandrov’s lemma Alexandrov’s lemma is a lemma that enables us to glue CAT(0) space together to obtain new examples. Lemma (Alexandrov’s lemma). Suppose the triangles ∆1 = ∆(x, y, z1) and ∆2 = ∆(x, y, z2) in a metric space satisfy the CAT(0) condition, and y ∈ [z1, z2]. x z1 y z2 Then ∆ = ∆(x, z1, z2) also satisﬁes the CAT(0) condition. This is the basic result we need if we want to prove “gluing theorems” for CAT(0) spaces. Proof. Consider ¯∆1 and ¯∆2, which together form a Euclidean quadrilateral ¯Q with with vertices ¯x, ¯z1, ¯z2, ¯y. We claim that then the interior angle at ¯y is ≥ 180◦. Suppose not, and it looked like this: 44 5 CAT(0) spaces and groups IV Geometric Group Theory ¯x ¯z1 ¯z2 ¯y If not, there exists ¯pi ∈ [¯y, ¯zi] such that [¯p1, ¯p2] ∩ [¯x, ¯y] = {¯q} and ¯q = ¯y. Now d(p1, p2) = d(p2, y) + d(y, p2) = d(¯p1, ¯y) + d(¯y, ¯p1) = d(¯p, ¯y) + d(¯y, ¯p2) > d(¯p1, ¯q) + d(¯q, ¯p2) ≥ d(p1, q) + d(q, p2) ≥ d(p1, p2), which is a contradiction. Thus, we know the right picture looks like this: ¯x ¯y ¯z1 ¯z2 To obtain ¯∆, we have to “push” ¯y out so that the edge ¯z1 ¯z2 is straight, while keeping the lengths ﬁxed. There is a natural map π : ¯∆ → ¯Q, and the lemma follows by checking that for any a, b ∈ ¯∆, we have d(π(a), π(b)) ≤ d(a, b). This is an easy case analysis (or is obvious). A sample application is the following: Proposition. If X1, X2 are both locally compact, complete CAT(0) spaces and Y is isometric to closed, subspaces of both X1 and X2. Then X1 ∪Y X2, equipped with the induced length metric, is CAT(0). 5.5 Cartan–Hadamard theorem Theorem (Cartan–Hadamard theorem). If X is a complete, connected length space of non-positive curvature, then the universal cover ˜X, equipped with the induced length metric, is CAT(0). This was proved by Cartan and Hadamard in the diﬀerential geometric setting. 45 5 CAT(0) spaces and groups IV Geometric Group Theory Corollary. A (torsion free) group Γ is CAT(0) iﬀ it is the π1 of a complete, connected space X of non-positive curvature. We’ll indicate some steps in the proof of the theorem. Lemma. If X is proper, non-positively curved and uniquely geodesic, then X is CAT(0). Proof idea. The idea is that given a triangle, we cut it up into a lot of small triangles, and since X is locally CAT(0), we can use Alexandrov’s lemma to conclude that the large triangle is CAT(0). Recall that geodesics vary continuously with their endpoints. Consider a triangle ∆ = ∆(x, y, z) ⊆ ¯B ⊆ X, where ¯B is a compact ball. By compactness, there is an ε such that for every x ∈ ¯B, the ball Bx(¯ε) is CAT(0). We let βt be the geodesic from x to α(t). Using continuity, we can choose 0 < t1 < · · · < tN = 1 such that d(βti(s), βti+1(s)) < ε for all s ∈ [0, 1]. Now divide ∆ up into a “patchwork” of triangles, each contained in an ε ball, so each satisﬁes the CAT(0) condition, and apply induction and Alexandrov’s lemma to conclude. Now to prove the Cartan–Hadamard theorem, we only have to show that the universal cover is uniquely geodesic. Here we must use the simply-connectedness condition. Theorem. Let X be a proper length space of non-positive curvature, and p, q ∈ X. Then each homotopy class of paths from p to q contains a unique (local) geodesic representative. 5.6 Gromov’s link condition Gromov’s link condition is a criterion that makes it very easy to write down interesting examples of non-positively-curved metric spaces. Deﬁnition (Euclidean cell complex). A locally ﬁnite cell complex X is Euclidean if every cell is isometric to a convex polyhedron in Euclidean space and the attaching maps are isometries from the lower-dimensional cell to a face of the new cell. Such a complex X has a natural length metric which is proper and geodesic by Hopf–Rinow. What we’d like to do is to come up with a condition that ensures X is CAT(0). Example. The usual diagram for a torus gives an example of a Euclidean complex. 46 5 CAT(0) spaces and groups IV Geometric Group Theory Example. We can construct a sphere this way, just by taking a cube! We know that T 2 is non-positively curved (since it is ﬂat), but the cube is not, because by Cartan–Hadamard, it would be CAT(0), hence contractible, but it is not. Deﬁnition (Link). Let X be a Euclidean complex, and let v be a vertex of X, and let 0 < ε shortest 1-cell. Then the link of v is Lk(v) = Sv(ε) = {x ∈ X : d(x, v) = ε}. Note that Lk(V ) is a cell complex: the intersection of Lk(v) with a cell of X of dimension is a cell of dimension n − 1. Example. In the torus, there is only one vertex. The link looks like So the link is S1. Example. If we take the corner of a cube, then Lk(v) is homeomorphic to S1, but it is a weird one, because it is made up of three right angles, not two. How can we distinguish between these two S1’s? Angle puts a metric on Lk(v). We can do this for general metric spaces, but we only need it for Euclidean complexes, in which case there is not much to do. Restricted to each cell, the link is just a part of a sphere, and it has a natural spherical metric, which is a length metric. These then glue together to a length metric on Lk(v). Note that this is not the same as the induced metric. Example. In the torus, the total length of the link is 2π, while that of the cube is 3π 2 . The important theorem, which we are not going to prove, is this: Theorem (Gromov’s link criterion). A Euclidean complex X is non-positivelycurved iﬀ for every vertex v of X, Lk(v) is CAT(1). Note that by deﬁnition, we only have to check the CAT(1) inequality on triangles of perimeter < 2π. 47 5 CAT(0) spaces and groups IV Geometric Group Theory Exercise. Check these in the case of the torus and the cube. Thus, given a group, we can try to construct a space whose π1 is it, and then put a Euclidean structure on it, then check Gromov’s link criterion. In general, Gromov’s link condition might not be too useful, because we still don’t know how to check if something is CAT(1)! However, it is very simple in dimension 2. Corollary. If X is a 2-dimensional Euclidean complex, then for all vertices v, Lk(v) is a metric graph, and X is CAT(0) iﬀ Lk(v) has no loop of length < 2π for all v. Example (Wise’s example). Consider the group W = a, b, s, t \| [a, b] = 1, s−1as = (ab)2, t−1bt = (ab)2. Letting c = ab, it is easy to see that this is the π1 of the following Euclidean complex: c a b s t This is metrized in the obvious way, where all edges have length 2 except the black ones of length 1. To understand the links, we set α = sin−1 1 4 , β = cos−1 1 4 . Then the triangles each has angles 2α, β, β. We show that W is non-positively curved, and then use this to show that there is a homomorphism W → W that is surjective but not injective. We say W is non-Hopﬁan. In particular, this will show that W is not linear, i.e. it is not a matrix group. To show that X is non-positively curved, we check the link condition To check the link condition, we have to check that there are no loops of length < 2π in Lk(v). Note that by trigonometric identities, we know α + β = π 2 . So we can just observe that everything is ﬁne. 48 5 CAT(0) spaces and groups IV Geometric Group Theory To see that W is non-Hopﬁan, we deﬁne f : W → W a → a2 b → b2 s → s2 t → t We check that this is a well-deﬁned homomorphism, and is surjective, since we observe a = sa2b2s−1, b = ta2b2t−1, and so a, b, s, t ∈ im t. The non-trivial claim is that ker f = 0. Let g = [scs−1, tct−1]. Note that f (g) = [f (scs−1), f (tct−1)] = [sc2s−1, tc2t−1] = [a, b] = 1. So the crucial claim is that g = 1. This is where geometry comes to the rescue. For convenience, write p = scs−1, q = tct−1. Consider the following local geodesics in the two squares: c s t The left- and right-hand local geodesics represent p and q respectively. Then g = p · q · ¯p · ¯q. We claim that this represents g by a local geodesic. The only place that can go wrong is the points where they are joined. By the proof of the Gromov link condition, to check this, we check that the three “turns” at the vertex are all of angle ≥ π. c a b p− q− q+ p+ b a c t s s t 49 5 CAT(0) spaces and groups IV Geometric Group Theory Here p+ is the point where p ends and p− is the point where p starts, and similarly for q. Moreover, the distance from p±, q± to the top/bottom left/right vertices is β. So we can just check and see that everything works. Recall that every homotopy class of paths in X contains a unique locally geodesic representative. Since the constant path is locally geodesic, we know g = 1. Deﬁnition (Residually ﬁnite group). A group G is residually ﬁnite if for every g ∈ G \ {1}, there is a homomorphism ϕ : G → ﬁnite group such that ϕ(g) = 0. Theorem (Mal’cev). Every ﬁnitely generated linear subgroup (i.e. a subgroup of GLn(C)) is resudially ﬁnite. Proof sketch. If the group is in fact a subgroup of GLn(Z), then we just reduce mod p for p 0. To make it work over GLn(C), we need a suitable version of the Nullstellensatz. Theorem (Mal’cev). Every ﬁnitely generated residually ﬁnite group is Hopﬁan. Proof. Finding a proof is a fun exercise! We know that Wise’s example is not Hopﬁan, hence not residually ﬁnite, hence not a linear group. Contrast this with the amazing theorem of Sela that all hyperbolic groups are Hopﬁan. However, a major open question is whether all hyperbolic groups are residually ﬁnite. On the other hand, it is known that not all hyperbolic groups are not linear. How are we supposed to think about residually ﬁnite groups? Lemma (Scott’s criterion). Let X be a cell complex, and G = π1X. Then G is residually ﬁnite if and only if the following holds: Let p : ˜X → X be the universal cover. For all compact subcomplexes K
⊆ ˜X, there is a ﬁnite-sheeted cover X → X such that the natural covering map p : ˜X → X is injective on K. A good (though not technically correct) way to think about this is follows: if we have a map f : K → X that may be complicated, and in particular is not injective, then we might hope that there is some “ﬁnite resolution” X → X such that f lifts to X , and the lift is injective. Of course, this is not always possible, and a necessary condition for this to work is that the lift to the universal cover is injective. If this is not true, then obviously no such resolution can exist. And residual ﬁniteness says if there is no obvious reason to fail, then it in fact does not fail. 5.7 Cube complexes Deﬁnition (Cube complex). A Euclidean complex is a cube complex if every cell is isometric to a cube (of any dimension). This is less general than Euclidean complexes, but we can still make high dimension things out of this. Of course, general Euclidean complexes also work in high dimensions, but except in two dimensions, the link condition is rather tricky to check. It turns out the link condition is easy to check for cube complexes. 50 5 CAT(0) spaces and groups IV Geometric Group Theory Purely topologically, the link of each vertex is made out of simplices, and, subdividing if necessary, they are simplicial complexes. The metric is such that every edge has length π 2 , and this is called the “all-right simplicial complex”. Recall that is non-positively curved, while is not. Deﬁnition (Flag simplicial complex). A simplicial complex X is ﬂag if for all n ≥ 2, each copy of ∂∆n in X is in fact the boundary of a copy of ∆n in X. Note that topologically, ﬂag complexes are not special in any sense. For any simplicial complex K, the ﬁrst barycentric subdivision is ﬂag. Theorem (Gromov). A cube complex is non-positively curved iﬀ every link is ﬂag. Now the property of being ﬂag is purely combinatorial, and easy to check. So this lets us work with cube complexes. Right-angled Artin groups Deﬁnition (right-angled Artin group). Let N be a simplicial graph, i.e. a graph where the vertices determine the edges, i.e. a graph as a graph theorist would consider. Then AN = V (N ) \| [u, v] = 1 for all (u, v) ∈ E(N ) is the right-angled Artin group, or graph group of N . Example. If N is the discrete graph on n vertices, then AN = Fn. Example. If N is the complete graph on n vertices, then AN = Zn. 51 5 CAT(0) spaces and groups IV Geometric Group Theory Example. If N is a square, i.e. the complete bipartite graph K2,2, then AN = F2 × F2. Example. When N is the path with 4 vertices, then this is a complicated group that doesn’t have a good, alternative description. This is quite an interesting group. Deﬁnition (Salvetti complex). Given a simplicial group N , the Salvetti complex SN is the cube complex deﬁned as follows: – Set S (2) N is the presentation complex for AN . – For any immersion of the 2-skeleton of a d-dimensional cube, we glue in an d-dimensional cube to S (2) N . Alternatively, we have a natural inclusion S (2) largest subcomplex whose 2-skeleton coincides with S (2) N . N ⊆ (S1)\|V (N )\|, and SN is the Example. If N is then An = Z ∗ Z2, and SN is a circle glued to a torus. Example. If N is , then SN is T 2 ∨ S1. There is a unique vertex v, and its link looks like In fact, there is a recipe for getting the link out of N . Deﬁnition (Double). The double D(K) of a simplicial complex K is deﬁned as follows: – The vertices are {v+ 1 , . . . , v+ n , v− 1 , . . . , v− n }, where {v1, . . . , vn} are the ver- tices of k. – The simplices are those of the form v± i0 , . . . , v± ik , where vi0 , . . . , vik ∈ K. Example. In the example above, the double of N is just Lk(v)! Deﬁnition (Flag complex). The ﬂag complex of N , written ¯N , is the only ﬂag simplicial complex with 1-skeleton N . Lemma. For any (simplicial) graph N , the link of the unique vertex of SN is D( ¯N ). In particular, SN is non-positively curved. Thus, right-angled Artin groups and their Salvetti complexes give examples of non-positively curved spaces with very general links. It turns out Theorem. Right-angled Artin groups embed into GLnZ (where n depends on N ). 52 5 CAT(0) spaces and groups IV Geometric Group Theory 5.8 Special cube complexes Let X be a non-positively curved cube complex. We will write down explicit geometric/combinatorial conditions on X so that π1X embeds into AN for some N . Hyperplanes and their pathologies If C ∼= [−1, 1]n, then a midcube M ⊆ C is the intersection of C with {xi = 0} for some i. Now if X is a non-positively curved cube complex, and M1, M2 are midcubes of cubes in X, we say M1 ∼ M2 if they have a common face, and extend this to an equivalence relation. The equivalence classes are immersed hyperplanes. We usually visualize these as the union of all the midcubes in the equivalence class. Note that in general, these immersed hyperplanes can have self-intersections, hence the word “immersed”. Thus, an immersed hyperplane can be thought of as a locally isometric map H X, where H is a cube complex. In general, these immersed hyperplanes can have several “pathologies”. The ﬁrst is self-intersections, as we have already met. The next problem is that of orientation, or sidedness. For example, we can have the following cube complex with an immersed hyperplane: This is bad, for the reason that if we think of this as a (−1, 1)-bundle over H, then it is non-orientable, and in particular, non-trivial. In general, there could be self intersections. So we let NH be the pullback interval bundle over H. That is, NH is obtained by gluing together {M ×(−1, 1) \| M is a cube in H}. Then we say H is two-sided if this bundle is trivial, and one-sided otherwise. Sometimes, we might not have self-intersections, but something like this: 53 5 CAT(0) spaces and groups IV Geometric Group Theory This is a direct self-osculation. We can also have indirect osculations that look like this: Note that it makes sense to distinguish between direct and indirect osculations only if our hyperplane is two-sided. Finally, we have inter-osculations, which look roughly like this: Haglund and Wise deﬁned special cube complexes to be cube complexes that are free of pathologies. Deﬁnition (Special cube complex). A cube complex is special if its hyperplanes do not exhibit any of the following four pathologies: – One-sidedness – Self-intersection – Direct self-osculation – Inter-osculation Example. A cube is a special cube complex. Example. Traditionally, the way to exhibit a surface as a cube complex is to ﬁrst tile it by right-angled polygons, so that every vertex has degree 4, and then the dual exhibits the surface as a cube complex. The advantage of this approach is that the hyperplanes are exactly the edges in the original tiling! From this, one checks that we in fact have a special curve complex. This is one example, but it is quite nice to have inﬁnitely many. It is true that covers of special things are special. So this already gives us inﬁnitely many special cube complexes. But we want others. Example. If X = SN is a Salvetti complex, then it is a special cube complex, and it is not terribly diﬃcult to check. The key theorem is the following: Theorem (Haglund–Wise). If X is a compact special cube complex, then there exists a graph N and a local isometry of cube complexes Corollary. π1X → AN . ϕX : X SN . 54 5 CAT(0) spaces and groups IV Geometric Group Theory Proof of corollary. If g ∈ π1X, then g is uniquely represented by a local geodesic γ : I → X. Then ϕ ◦ γ is a local geodesic in SN . Since homotopy classes of loops are represented by unique local geodesics, this implies ϕX ◦ γ is not null-homotopic. So the map (ϕX )∗ is injective. So if we know some nice group-theoretic facts about right-angled Artin groups, then we can use them to understand π1X. For example, Corollary. If X is a special cube complex, then π1X is linear, residually ﬁnite, Hopﬁan, etc. We shall try to give an indication of how we can prove the Haglund–Wise theorem. We ﬁrst make the following deﬁnition. Deﬁnition (Virtually special group). A group Γ is virtually special if there ∼= π1X, where X is a compact exists a ﬁnite index subgroup Γ0 ≤ Γ such that Γ0 special cube complex. Sketch proof of Haglund–Wise. We have to ﬁrst come up with an N . We set the vertices of N to be the hyperplanes of X, and we join two vertices iﬀ the hyperplanes cross in X. This gives SN . We choose a transverse orientation on each hyperplane of X. Now we deﬁne ϕX : X SN cell by cell. – Vertices: There is only one vertex in SN . – Edges: Let e be an edge of X. Then e crosses a unique hyperplane H. Then H is a vertex of N . This corresponds to a generator in AN , hence a corresponding edge e(H) of SN . Send e to e(H). The choice of transverse orientation tells us which way round to do it – Squares: given hyperplanes e1 f1 H H f2 e2 Note that we already mapped e1, e2 to e(H), and f1, f2 to e(H ). Since H and H cross in X, we know e(H) and e(H ) bound a square in SN . Send this square in X to that square in SN . – There is nothing to do for the higher-dimensional cubes, since by deﬁnition of SN , they have all the higher-dimensional cubes we can hope for. We haven’t used a lot of the nice properties of special cube complexes. They are needed to show that the map is a local isometric embedding. What we do is to use the hypothesis to show that the induced map on links is an isometric embedding, which implies ϕX is a local isometry. 55 5 CAT(0) spaces and groups IV Geometric Group Theory This really applies to a really wide selection of objects. Example. The following groups are virtually special groups: – π1M for M almost any 3-manifold. – Random groups This is pretty amazing. A “random group” is linear! 56 Index Index Z, 23 BG, 25 D(K), 52 F (S), 9 Fr, 9 K(G, 1), 25 X Z XP , 11 FArea, 22 δ-hyperbolic space, 29 δ-slim triangle, 29 ∂∞X, 39 , 15 dHaus, 30 k-local geodesic, 34 Alexandrov’s lemma, 44 algebraic area, 14 asph
erical space, 25 Bass–Serre tree, 27 Britton’s lemma, 28 Cartan–Hadamard theorem, 40, 45 CAT(κ) space, 41 CAT(0) group, 43 Cayley graph, 3 cocompact action, 7 comparison point, 41 comparison triangle, 41 cube complex, 50 special, 54 curvature, 41 cyclicly reduced word, 34 Dehn function, 14 Dehn presentation, 37 direct self-osculation, 54 disc diagram, 18 dissection, 31 double, 52 Eilenberg–MacLane spaces, 25 elementary reductions, 13 Euclidean cell complex, 46 face labels, 18 ﬁlling disc, 22 ﬁnitely-presentable group, 11 IV Geometric Group Theory ﬁnitely-presented group, 11 ﬂag complex, 52 ﬂag simplicial complex, 51 free group, 9 geodesic, 6 geodesic metric space, 6 geodesic ray, 39 geodesic triangle, 29 graph of groups, 25 of spaces, 23 graph group, 51 Gromov hyperbolic space, 29 Gromov’s link criterion, 47 group cohomology, 40 group homology, 40 Hausdorﬀ distance, 30 HNN extension, 24 homology sphere, 12 homotopy pushout, 23 Hopf–Rinow theorem, 7, 44 hyperbolic group, 31 hyperbolic plane, 29 hyperbolic space, 29 indirect osculations, 54 inter-osculations, 54 isoperimetric function, 22 length of path, 31 length pseudometric, 44 length space, 44 link, 47 local geodesic, 34 metric space geodesic, 6 midcube, 53 Morse lemma, 30 non-Hopﬁan, 48 non-positively curved space, 41 normal form theorem, 28 Novikov–Boone theorem, 21, 40 null-homotopic, 14 one-sided, 53 57 Index IV Geometric Group Theory Poincar´e duality, 40 Poincar´e homology sphere, 12 presentation, 11 proper discontinuous action, 7 proper metric space, 6 quasi-convex, 38 quasi-geodesic, 30 quasi-geodesic interval, 30 quasi-geodesic line, 30 quasi-geodesic ray, 30 quasi-injective, 5 quasi-isometric, 5 quasi-isometric embedding, 5 quasi-isometry, 5 quasi-surjective, 5 random group, 34 realization, 24 rectiﬁable path, 31 recursively enumerable, 13 reduced word, 10, 13 residually ﬁnite group, 50 right-angled Artin group, 51 rose with r petals, 9 Salvetti complex, 52 Scott’s criterion, 50 Seifert–van Kampen theorem, 11 singular disc diagram, 18 special cube complex, 54 triangle, 41 geodesic, 29 tripod, 29 two-sided, 53 van Kampen diagram, 18 van Kampen’s lemma, 19 virtually special proof, 55 virtually-P, 34 Wise’s example, 48 word length, 3 word metric, 3 58
an view K as the Q-points of an n-dimensional aﬃne group scheme. We can then deﬁne RK/QGm ⊆ A to be the set on which X is invertible, and then an algebraic Hecke character is a homomorphism of algebraic groups (TK)/C → Gm/C, where TK = ResK/Q(Gm). If we have a real place v of K, then this corresponds to a real embedding ∼= R, and if v is a complex place, we have a pair of embedding σv : K → Kv σv, ¯σv : K → Kv C, picking one of the pair to be σv. So ϕ extends to a homomorphism ϕ : K × ∞ → C× given by ϕ(xv) = xn(σv) v xn(σv) v ¯xn(¯σv) v v real v complex 23 2 L-functions IV Topics in Number Theory Deﬁnition (Algebraic Hecke character). A Hecke character χ = χ∞χ∞ : JK/K × → C× is algebraic if there exists an algebraic homomorphism ϕ : K × → C× such that ϕ(x) = χ∞(x) for all x ∈ K ×,0 v real sgnev v for ev ∈ {0, 1}. ∞ , i.e. χ∞ = ϕ We say ϕ (or the tuple (n(σ))σ) is the inﬁnite type of χ. Example. The adelic norm \| · \|A : JK → C× has χ∞ = \| · \|v, and so χ is algebraic, and the associated ϕ is just NK/Q : K × → Q× ⊆ C×, with (nσ) = (1, . . . , 1). Exercise. Let K = Q(i), and χ from the previous example, whose associated character of ideals was Θ : p → πp, where πp ≡ 1 mod (2 + 2i). The inﬁnity type is the inclusion K × → C×, i.e. it has type (1, 0). Observe that the image of an algebraic homomorphism ϕ : K × → C× lies in the normal closure of K. More generally, Proposition. If χ is an algebraic Hecke character, then χ∞ takes values in some number ﬁeld. We write E = E(χ) for the smallest such ﬁeld. Of course, we cannot expect χ to take algebraic values, since JK contains copies of R and C. Proof. Observe that χ∞( ˆO× totally positive. Then K) is ﬁnite subgroup, so is µn for some n. Let x ∈ K ×, χ∞(x) = χ∞(x)−1 = ϕ(x)−1 ∈ K cl, where K cl is the Galois closure. Then since K × (by the ﬁniteness of the class group), so >0 × ˆO× K → ˆK × has ﬁnite cokernel χ∞( ˆK ×) = d i=1 ziχ∞(K × >0 ˆO× K), i ∈ χ∞(K × >0 where zd the image of K × ˆO× K), and is therefore contained inside a ﬁnite extension of >0 × ˆO× K. Hecke characters of ﬁnite order (i.e. algebraic Hecke characters with inﬁnity type (0, . . . , 0)) are in bijection with continuous homomorphisms ΓK → C×, necessarily of ﬁnite order. What we show now is how to associate to a general algebraic Hecke character χ a continuous homomorphism ψ : ΓK → E(χ)× λ ⊇ Q× , where λ is a place of E(χ) over . This is continuous for the -adic topology on Eλ. In general, this will not be of ﬁnite order. Thus, algebraic Hecke characters correspond to -adic Galois representations. The construction works as follows: since χ∞(x) = ϕ(x)−1, we can restrict the inﬁnity type ϕ to a homomorphism ϕ : K × → E×. We deﬁne ˜χ : JK → E× as follows: if x = x∞x∞ ∈ K × ∞K∞,× ∈ JK, then we set ˜χ(x) = χ(x)ϕ(x∞)−1. 24 2 L-functions IV Topics in Number Theory Notice that this is not trivial on K × in general. Then ˜χ∞ takes values in {±1}. Thus, ˜χ takes values in E×. Thus, we know that ˜χ has open kernel, i.e. it is continuous for the discrete topology on E×, and ˜χ\|K× = ϕ−1. Conversely, if ˜χ : K × → E× is a continuous homomorphism for the discrete topology on E×, and ˜χ\|K× is an algebraic homomorphism, then it comes from an algebraic Hecke character in this way. Let λ be a ﬁnite place of E over , a rational prime. Recall that ϕ : K × → E× is an algebraic homomorphism, i.e. xiei ϕ = f (x), f ∈ E(X1, . . . , Xn). We can extend this to K × = (K ⊗Q Q)× = v\| K × v to get a homomorphism ϕλ : K × → E× λ This is still algebraic, so it is certainly continuous for the -adic topology. Now consider the character ψλ : JK → E× λ , where now ψλ((xv)) = ˜χ(x)ϕλ((xv)v\|). λ , and moreover, we see ∞ ) = {1}, we This is then continuous for the -adic topology on E× that ψλ(K ×) = {1} as ˜χ\|K× = ϕ−1 while ϕλ\|K× = ϕ. Since ˜χ(K ×,0 know that ψλ it is in fact deﬁned on CK/C 0 K Obviously, ψλ determines ˜χ and hence χ. K → E× Fact. An -adic character ψ : CK/C 0 λ comes from an algebraic Hecke character in this way if and only if the associated Galois representation is Hodge–Tate, which is a condition on the restriction to the decomposition groups Gal( ¯Kv/Kv) for the primes v \| . Example. Let K = Q and χ = \| · \|A, then ∼= Γab K . ˜χ = sgn(x∞) p \|xp\|p. So ψ((xv)) = sgn(x∞) p= \|xp\|p · \|x\| · x. Note that \|x\|x ∈ Z× . We have CQ/C 0 Q ∼= ˆZ×. Under this isomorphism, the map ˆZ×Q× by the inclusion, and by class ﬁeld theory, ψ : Gal( ¯Q/Q) → Z× cyclotomic character of the ﬁeld Q({ζn }), is just the projection onto Z× followed is just the σ(ζn ) = ζ ψ(σ) mod n n . Example. Consider the elliptic curve y2 = x3 − x with complex multiplication over Q(i). In other words, End(E/Q(i)) = Z[i], where we let i act by i · (x, y) → (−x, iy). 25 2 L-functions IV Topics in Number Theory Its Tate module is a Z[i]-module. If λ \| , then we deﬁne TE = lim E[n] VλE = TE ⊗Z[i] Kλ. Then ΓK act by ΓK : AutKλ VλE = K × λ . We now want to study the inﬁnity types of an algebraic Hecke character. Lemma. Let K be a number ﬁeld, ϕ : K × → E× ⊆ C× be an algebraic homomorphism, and suppose E/Q is Galois. Then ϕ factors as K × norm−→ (K ∩ E)× φ −→ E×. Note that since E is Galois, the intersection K ∩ E makes perfect sense. Proof. By deﬁnition, we can write ϕ(x) = σ(x)n(σ). σ:K→C Then since ϕ(x) ∈ E, for all x ∈ K × and τ ∈ ΓE, we have τ σ(x)n(σ) = σ(x)n(σ). In other words, we have σ(x)n(τ −1σ) = σ σ(x)n(σ). σ Since the homomorphisms σ are independent, we must have n(τ σ) = n(σ) for all embeddings σ : K → ¯Q and τ ∈ ΓE. This implies the theorem. Recall that if m is a modulus, then we deﬁned open subgroups Um ⊆ JK, consisting of the elements (xv) such that if a real v \| m, then xv > 0, and if v \| m for a ﬁnite v, then v(xv − 1) ≥ mv. We can write this as Um = Um,∞ × U ∞ m . Proposition. Let ϕ : K × → C× be an algebraic homomorphism. Then ϕ is the inﬁnity type of an algebraic Hecke character χ iﬀ ϕ(O× K) is ﬁnite. Proof. To prove the (⇒) direction, suppose χ = χ∞χ∞ is an algebraic Hecke character with inﬁnity type ϕ. Then χ∞(U ∞ m ) = 1 for some m. Let Em = K × ∩ Um ⊆ O× K, a subgroup of ﬁnite index. As χ∞(Em) = 1 = χ(Em), we know χ∞(Em) = 1. So ϕ(O× K) is ﬁnite. To prove (⇐), given ϕ with ϕ(O× ϕ(Em) = 1. Then (ϕ, 1) : K × this to a homomorphism K) ﬁnite, we can ﬁnd some m such that m → C× is trivial on Em. So we can extend ∞ × U ∞ K × ∞UmK × K × ∼= K × ∞Um Em → C×, since Em = K × ∩ Um. But the LHS is a ﬁnite index subgroup of CK. So the map extends to some χ. 26 2 L-functions IV Topics in Number Theory Here are some non-standard terminology: Deﬁnition (Serre type). A homomorphism ϕ : K × → C× is of Serre type if it is algebraic and ϕ(O× K) is ﬁnite. These are precisely homomorphisms that occur as inﬁnity types of algebraic Hecke characters. Note that the unit theorem implies that O× K → K ×,1 ∞ = {x ∈ K × ∞ : \|x\|A = 1} has compact cokernel. If ϕ(O× into U(1). K) is ﬁnite, then ϕ(K ×,1 ∞ ) is compact. So it maps Example. Suppose K is totally real. Then Then we have K × ∞ = (R×){σ:K→R}. K ×,1 ∞ = {(xσ) : xσ = ±1}. Then ϕ((xσ)) = xn(σ) σ Thus, ϕ is just a power of the norm map. , so \|ϕ(K ×,1 ∞ )\| = 1. In other words, all the nσ are equal. Thus, algebraic Hecke characters are all of the form \| · \|m A · (ﬁnite order character). Another class of examples comes from CM ﬁelds. Deﬁnition (CM ﬁeld). K is a CM ﬁeld if K is a totally complex quadratic extension of a totally real number ﬁeld K +. This CM refers to complex multiplication. This is a rather restrictive condition, since this implies Gal(K/K +) = {1, c} = v ) for every w \| v \| ∞. So c is equal to complex conjugation for every Gal(Kw/K + embedding K → C. From this, it is easy to see that CM ﬁelds are all contained in QCM ⊆ ¯Q ⊆ C, given by the ﬁxed ﬁeld of the subgroup cσcσ−1 : σ ∈ ΓQ ⊆ ΓQ. For example, we see that the compositum of two CM ﬁelds is another CM ﬁeld. Exercise. Let K be a totally complex S3-extension over Q. Then K is not CM, but the quadratic subﬁelds is complex and is equal to K ∩ QCM. Example. Let K be a CM ﬁeld of degree 2r. Then Dirichlet’s unit theorem tells us rk O× K = r − 1 = rk O× K+. K is a ﬁnite index subgroup of O× K+. So ϕ : K × → C× is of Serre type iﬀ So O× it is algebraic and its restriction to K +,× is of Serre type. In other words, we need n(σ) + n(¯σ) to be independent of σ. 27 2 L-functions IV Topics in Number Theory Theorem. Suppose K is arbitrary, and ϕ : K × → E× ⊆ C× is algebraic, and we assume E/Q is Galois, containing the normal closure of K. Thus, we can write ϕ(x) = σ(x)n(σ). Then the following are equivalent: (i) ϕ is of Serre type. σ:K→E (ii) ϕ = ψ ◦ NK/F , where F is the maximal CM subﬁeld and ψ is of Serre type. (iii) For all c ∈ Gal(E/Q) conjugate to complex conjugation c, the map σ → n(σ) + n(cσ) is constant. (iv) (in the case K ⊆ C and K/Q is Galois with Galois group G) Let λ = n(σ)σ ∈ Z[G]. Then for all τ ∈ G, we have (τ − 1)(c + 1)λ = 0 = (c + 1)(τ − 1)λ. Note that in (iii), the constant is necessarily 2 [K : Q] σ n(σ). So in particular, it is independent of c. Proof. – (iii) ⇔ (iv): This is just some formal symbol manipulation. – (ii) ⇒ (i): The norm takes units to units. – (i) ⇒ (iii): By the previous lecture, we know that if ϕ is of Serre type, then \|ϕ(K ×,1 ∞ )\| = 1. Now if (xv) ∈ K × \|ϕ((xv))\| = ∞, we have \|xv\|n(σv) \|xv\|n(σv)+n(¯σv) = 1 2 (n(σv)+n(¯σv)) \|xv\| v . real v complex v v Here the modulus without the subscript is the usual modulus. Then \|ϕ(K ×,1 ∞ )\| = 1 implies n(σv) + n(¯σv) is constant. In other words, n(σ) + n(cσ) = m is constant. But if τ ∈ Gal(E/Q), and ϕ = τ ◦ ϕ, n(σ) = n(τ −1σ), then this is also of Serre type. So m = n(σ) + n(cσ) = n(τ −1σ) + n(τ −1cσ) = n(τ −1σ) + n((τ −1cτ )τ −1σ). – (iii) ⇒ (ii): Suppose n(σ) + n(cσ) = m for all σ and all c = τ cτ −1. Then we must have n(cσ) = n(cσ) 28 2 L-functions IV Topics in Number Theory for all σ. So n(σ) = n(cτ cτ −1σ) So n is invariant under H = [c, Gal(E/Q)] ≤ Gal(E/Q), noting that c has order 2. So ϕ takes values in the ﬁxed ﬁeld EH = E ∩ QCM. By the proposition last time, this implies ϕ factors thr
ough NK/F , where F = EH ∩ K = K ∩ QCM. Recall that a homomorphism ϕ : K × → C× is algebraic iﬀ it is a character of the commutative algebraic group TK = RK/QGm, so that TK(Q) = K ×, i.e. there is an algebraic character ϕ : TK/C → Gm/C such that ϕ restricted to TK(Q) is ϕ. Then ϕ is of Serre type iﬀ ϕ is a character of KS0 = TK/E 0 K, where EK is the K is the identity component, which is the same K, where ∆ is a suﬃciently small ﬁnite-index Zariski closure of O× K in TK and E 0 as the Zariski closure of ∆ ⊆ O× subgroup. The group KS0 is called the connected Serre group. We have a commutative diagram (with exact rows) 1 1 K × KS0 JK KS JK/K × π0 Γab K 1 1 This KS is a projective limit of algebraic groups over Q. We have Hom(KS, C×) = Hom(KS, Gm/C) = {algebraic Hecke characters of K} The inﬁnity type is just the restriction to KS0. Langlands created a larger group, the Tamiyama group, an extension of Gal( ¯Q/Q) by KS0, which is useful for abelian varieties with CM and conjugations and Shimura varieties. 2.2 Abelian L-functions We are now going to deﬁne L-functions for Hecke characters. Recall that amongst all other things, an L-function is a function in a complex variable. Here we are going to do things slightly diﬀerently. For any Hecke character χ, we will deﬁne L(χ), which will be a number. We then deﬁne L(χ, s) = L(\| · \|s Aχ). We shall deﬁne L(χ) as an Euler product, and then later show it can be written as a sum. Deﬁnition (Hecke L-function). Let χ : CK → C× be a Hecke character. For v ∈ ΣK, we deﬁne local L-factors L(χv) as follows: – If v is non-Archimedean and χv unramiﬁed, i.e. χv\|O× Kv = 1, we set L(χv) = 1 1 − χv(πv) . 29 2 L-functions IV Topics in Number Theory – If v is non-Archimedean and χv is ramiﬁed, then we set L(χv) = 1. – If v is a real place, then χv is of the form χv(x) = x−N \|x\|s v, where N = 0, 1. We write L(χv) = ΓR(s) = π−s/2Γ(s/2). – If v is a complex place, then χv is of the form χv(x) = σ(x)−N \|x\|s v, where σ is an embedding of Kv into C and N ≥ 0. Then L(χv) = ΓC(s) = 2(2π)−sΓ(s) We then deﬁne L(χv, s) = L(χv · \| · \|s v). So for ﬁnite unramiﬁed v, we have L(χv, s) = 1 1 − χv(πv)q−s v , where qv = \|OKv /(πv)\|. Finally, we deﬁne L(χ, s) = L(χv, s) Λ(χ, s) = v∞ L(χv, s). v Recall that the kernel of the idelic norm \| · \|A : CK → R× >0 is compact. It is then not hard to see that for every χ, there is some t ∈ R such that χ · \| · \|t A is unitary. Thus, L(χ, s) converges absolutely on some right half-plane. Observe that Λ(χ\| · \|t A, s) = Λ(χ, t + s). Theorem (Hecke–Tate). (i) Λ(χ, s) has a meromorphic continuation to C, entire unless χ = \| · \|t some t ∈ C, in which case there are simple poles at s = 1 − t, −t. A for (ii) There is some function, the global ε-factor , ε(χ, s) = ABs for some A ∈ C× and B ∈ R>0 such that Λ(χ, s) = ε(χ, s)Λ(χ−1, 1 − s). 30 2 L-functions IV Topics in Number Theory (iii) There is a factorization ε(χ, s) = v εv(χv, µv, ψv, s), where εv = 1 for almost all v, and εv depends only on χv and certain auxiliary data ψv, µv. These are the local ε-factors. Traditionally, we write and then L(χ, s) = ﬁnite v L(χv, s), Λ(χ, s) = L(χ, s)L∞(χ, s). However, Tate (and others, especially the automorphic people) use L(χ, s) for the product over all v. At ﬁrst, Hecke proved (i) and (ii) using global methods, using certain Θ functions. Later, Tate proved (i) to (iii) using local-global methods and especially Fourier analysis on Kv and AK. This generalizes considerably, e.g. to automorphic representations. We can explain some ideas of Hecke’s method. We have a decomposition K∞ = K ⊗Q R ∼= Rr1 × Cr2 ∼= Cn, and this has a norm · induces by the Euclidean metric on Rn. Let ∆ ⊆ O× K,+ be a subgroup of totally positive units of ﬁnite index, which is ∼= Zr1+r2−1. This has an embedding ∆ = K ×,1 ∞ , which extends to a continuous homomorphism ∆ ⊗ R → K ×,1 ∞ . The key fact is Proposition. Let x ∈ K ×. Pick some invariant measure du on ∆ ⊗ R. Then ∆⊗R 1 ux2s du = stuﬀ \|NK/Q(x)\|2s/n , where the stuﬀ is some ratio of Γ factors and powers of π (and depends on s). Exercise. Prove this when K = Q[ √ d] for d > 0. Then ∆ = ε, and then ∞ LHS = 1 \|εtx + ε−tx\|2s dt RHS = −∞ stuﬀ \|xx\|s . The consequence of this is that if a ⊆ K is a fractional ideal, then 0=x∈a mod ∆ 1 \|NK/Q(x)\|s = stuﬀ · 0=x∈a mod ∆ ∆⊗R 1 uxns du = stuﬀ · ∆⊗R/∆   0=x∈a 1 uxns   du 31 2 L-functions IV Topics in Number Theory The integrand has a name, and is called the Epstein ζ-function of the lattice (a, u · 2). By the Poisson summation formula, we get an analytic continuity and functional equation for the epsilon ζ function. On the other hand, taking linear combinations of the left gives L(χ, s) for χ : Cl(K) → C×. For more general χ, we modify this with some extra factors. When the inﬁnity type is non-trivial, this is actually quite subtle. Note that if χ is unramiﬁed outside S and ramiﬁed at S, recall we had a homomorphism Θ : IS → C× sending pv → χv(πv)−1. So L(χ, s) = ﬁnite v∈S 1 1 − Θ(pv)−1(N pv)−s = a∈OK prime to S Θ(a)−1 (N a)s . This was Hecke’s original deﬁnition of the Hecke character. If K = Q and χ : CQ → C× is of ﬁnite order, then it factors through ∼= ˆZ× → (Z/N Z)×, and so χ is just some Dirichlet character CQ → CQ/C 0 Q ϕ : (Z/nZ)× → C×. The associated L-functions are just Dirichlet L-functions. Indeed, if p N , then χp(p) = χ(1, . . . , 1, p, 1, . . .) = χ(p−1, . . . , p−1, 1, p−1, . . .) = ϕ(p mod N )−1. In other words, L(χ, s) is the Dirichlet L-series of ϕ−1 (assuming N is chosen so that χ ramiﬁes exactly at v \| N ). Tate’s method uses local ε-factors ε(χv, µv, ψv, s), where ψv : Kv → U(1) is a non-trivial additive character, e.g. for v ﬁnite, KV tr Qp Qp/Zp ∼= Z[vp]/Z e2πix C×, which we needed because Fourier transforms take in additive measures, and µv is a Haar measure on Kv. The condition for (iii) to hold is ψv : AK → U(1) is well-deﬁned and trivial on K ⊆ AK, and µA = µv is a well-deﬁned measure on AK, i.e. µv(Ov) = 1 for all v and AK /K µA = 1. There exists explicit formulae for these εv’s. If χv is unramiﬁed, then it is just AvBs v, and is usually 1; for ramiﬁed ﬁnite v, they are given by Gauss sums. 2.3 Non-abelian L-functions Let K be a number ﬁeld. Then we have a reciprocity isomorphism ArtK : CK/C 0 K ∼→ Γab K . If χ : CK → C 0 as a map ψ = χ ◦ Art−1 K : ΓK → C×. Then K → C× is a Hecke character of ﬁnite order, then we can view it L(χ, s) = ﬁnite v unramiﬁed 1 1 − χv(πv)q−s v −1 = 1 1 − ψ(Frobv)q−s v , 32 2 L-functions IV Topics in Number Theory where Frobv ∈ ΓKv /IKv is the geometric Frobenius, using that ψ(IKv ) = 1. Artin generalized this to arbitrary complex representations of ΓK. Let ρ : ΓK → GLn(C) be a representation. Deﬁne L(ρ, s) = ﬁnite v L(ρv, s), where ρv is the restriction to the decomposition group at v, and depends only on the isomorphism class of ρ. We ﬁrst deﬁne these local factors for non-Archimedean ﬁelds: Deﬁnition. Let F be local and non-Archimedean. Let ρ : WF → GLC(V ) be a representation. Then we deﬁne L(ρ, s) = det(1 − q−sρ(FrobF )\|V IF )−1, where V IF is the invariants under IF . Note that in this section, all representations will be ﬁnite-dimensional and continuous for the complex topology (so in the case of WF , we require ker σ to be open). Proposition. (i) If is exact, then 0 → (ρ, V ) → (ρ, V ) → (ρ, V ) → 0 L(ρ, s) = L(ρ, s) · L(ρ, s). (ii) If E/F is ﬁnite separable, ρ : WE → GLC(V ) and σ = IndWF WE GLC(U ), then ρ : WF → L(ρ, s) = L(σ, s). Proof. (i) Since ρ has open kernel, we know ρ(IF ) is ﬁnite. So 0 → (V )IF → V IF → (V )IF → 0 is exact. Then the result follows from the multiplicativity of det. (ii) We can write U = {ϕ : WF → V : ϕ(gx) = ρ(g)ϕ(x) for all g ∈ WE, x ∈ WF }. where WF acts by Then we have σ(g)ϕ(x) = ϕ(xg). U IF = {ϕ : WF /IF → V : · · · }. Then whenever ϕ ∈ U IF and g ∈ IE, then σ(g)ϕ(x) = ϕ(xg) = ϕ((xgx−1)x) = ϕ(x). 33 2 L-functions IV Topics in Number Theory So in fact ϕ takes values in V IE . Therefore U IF = IndWF /IF WE /IE V IE . Of course, WF /IF ∼= Z, which contains WE/IE as a subgroup. Moreover, Frobd F = FrobE, where d = [kE : kF ]. We note the following lemma: Lemma. Let G = g ⊇ H = h = gd, ρ : H → GLC(V ) and σ = IndG Then H ρ. det(1 − tdρ(h)) = det(1 − tσ(g)). Proof. Both sides are multiplicative for exact sequences of representations of H. So we can reduce to the case of dim V = 1, where ρ(h) = λ ∈ C×. We then check it explicitly. To complete the proof of (ii), take g = FrobF and t = q−s F td = q−s E . so that For Archimedean F , we deﬁne L(ρ, s) in such a way to ensure that (i) and (ii) hold, and if dim V = 1, then L(ρ, s) = L(χ, s), where if ρ : W ab Artin map. F → C×, then χ Is the corresponding character of F × under the If F C, then this is rather easy, since every irreducible representation of ∼= C× is one-dimensional. We then just deﬁne for ρ 1-dimensional using ∼= F × and extend to all ρ by (i). The Jordan–H¨older theorem tells us this WF W ab F is well-deﬁned. If F R, then recall that WR = C×, s : s2 = −1 ∈ C×, szs−1 = ¯z. Contained in here is W (1) R = U(1), s. Then WR = W (1) R × R× >0. It is then easy to see that the irreducible representations of WR are (i) 1-dimensional ρWR ; or (ii) 2-dimensional, σ = IndWR C ρ, where ρ = ρs : C× → C×. In the ﬁrst case, we deﬁne L(ρ, s) = L(χ, s) using the Artin map, and in the second case, we deﬁne using (ii). L(σ, s) = L(ρ, s) 34 2 L-functions IV Topics in Number Theory To see that the properties are satisﬁed, note that (i) is true by construction, and there is only one case to check for (ii), which is if ρ = ρs, i.e. ρ(z) = (z ¯z)t. Then IndWR x → \|x\|t and x → sgn(x)\|x\|t = x−1\|x\|t+1. Then (ii) follows from the identity C× ρ is reducible, and is a sum of characters of W ab R ∼= R×, namely ΓR(s)ΓR(s + 1) = ΓC(s) = 2(2π)−sΓ(s). Now let K be global, and let ρ : ΓK → GLC(V ). For each v ∈ ΣK, choose ¯k of ∼= ΓKv be the decomposition group at ¯v. These contain IV , ¯K over v. Let Γv and we have the geometric Frobenius Frobv ∈ Γv/IV . We deﬁne ρv = ρ\|Γv , and then set L(ρ, s) = v∞ L(ρv, s)
= v∞ det(1 − q−s v Frobv\|V Iv )−1 Λ(ρ, s) = LL∞ L∞ = v\|∞ L(ρv, s). This is well-deﬁned as the decomposition groups ¯v \| v are conjugate. If dim V = 1, then ρ = χ ◦ Art−1 K for a ﬁnite-order Hecke character χ, and then The facts we had for local factors extend to global statements L(ρ, s) = L(χ, s). Proposition. (i) L(ρ ⊕ ρ, s) = L(ρ, s)L(ρ, s). (ii) If L/K is ﬁnite separable and ρ : ΓL → GLC(V ) and σ = IndΓK ΓL (ρ), then L(ρ, s) = L(σ, s). The same are true for Λ(ρ, s). Proof. (i) is clear. For (ii), we saw that if w ∈ ΣL over v ∈ ΣK and consider the local extension Lw/Kv, then L(ρw, s) = L(IndΓKv ΓLw ρw). In the global world, we have to take care of the splitting of primes. This boils down to the fact that IndΓK ΓL ρ ΓKv = w\|v IndΓKv ΓLw (ρ\|ΓLw ). (∗) We ﬁx a valuation ¯v of ¯K over v. Write Γ¯v/v for the decomposition group in ΓK. Write ¯S for the places of ¯K over v, and S the places of L over v. The Galois group acts transitively on ¯S, and we have ¯S ∼= ΓK/Γ¯v/v. 35 2 L-functions IV Topics in Number Theory We then have S ∼= ΓL\ΓK/Γ¯v/v, which is compatible with the obvious map ¯S → S. For ¯w = g¯v, we have Γ ¯w/v = gΓ¯v/vg−1. Conjugating by g−1, we can identify this with Γ¯v/v. Similarly, if w = ¯w\|L, then this contains and we can identify this with Γ¯v/v ∩ g−1ΓLg. Γ ¯w/w = gΓ¯v/vg−1 ∩ ΓL, There is a theorem, usually called Mackey’s formula, which says if H, K ⊆ G are two subgroups of ﬁnite index, and ρ : H → GLC(V ) is a representation of H. Then (IndG H V )\|K ∼= IndK K∩g−1Hg(g−1 V ), where g−1 then apply this to G = ΓK, H = ΓL, K = Γ¯v/v. V is the K ∩ g−1Hg-representation where g−1xg acts by ρ(x). We g∈H\G/K Example. If ρ is trivial, then L(ρ, s) = (1 − q−s v )−1 = This is just the Dedekind ζ-function of K. v aOK 1 N as = ζK(s). Example. Let L/K be a ﬁnite Galois extension with Galois group G. Consider the regular representation rL/K on C[G]. This decomposes as ρdi i , where {ρi} run over the irreducible representations of G of dimension di. We also have So by the induction formula, we have rL/K = IndΓK ΓL (1). ζL(s) = L(rL/K, s) = L(ρi, s)di. i Example. For example, if L/K = Q(ζN )/Q, then ζQ(ζN )(s) = L(χ, s), χ where the product runs over all primitive Dirichlet characters mod M \| N . Since ζQ(ζN ), ζQ have simple poles at s = 1, we know that L(χ, 1) = 0 if χ = χ0. Theorem (Brauer induction theorem). Suppose ρ : G → GLN (C) is a representation of a ﬁnite group. Then there exists subgroups Hj ⊆ G and homomorphisms χj : Hj → C× and integers mi ∈ Z such that tr ρ = j mj tr IndG Hj χj. Note that the mj need not be non-negative. So we cannot quite state this as a statement about representations. 36 2 L-functions IV Topics in Number Theory Corollary. Let ρ : ΓK → GLN (C). Then there exists ﬁnite separable Lj/K and χj : ΓLj → C× of ﬁnite order and mj ∈ Z such that L(ρ, s) = L(χj, s)mj . j In particular, L(ρ, s) has meromorphic continuation to C and has a functional equation Λ(ρ, s) = L · L∞ = ε(ρ, s)L(˜ρ, 1 − s) ε(ρ, s) = ABs = ε(χj, s)mj , where and ˜ρ(g) =t ρ(g−1). Conjecture (Artin conjecture). If ρ does not contain the trivial representation, then Λ(ρ, s) is entire. This is closely related to the global Langlands conjecture. In general, there is more than one way to write ρ as an sum of virtual induced characters. But when we take the product of the ε factors, it is always well-deﬁned. We also know that ε(χj, s) = εv(χj,v, s) is a product of local factors. It turns out the local decomposition is not independent of the decomposition, so if we want to write ε(ρ, s) = v εv(ρv, s), we cannot just take εv(ρv, s) = εv(χj,v, s), as this is not well-deﬁned. However, Langlands proved that there exists a unique factorization of ε(ρ, s) satisfying certain conditions. We ﬁx F a non-Archimedean local ﬁeld, χ : F × → C× and local ε factors ε(χ, ψ, µ), where µ is a Haar measure on F and ψ : F → U(1) is a non-trivial character. Let n(ψ) be the least integer such that ψ(πn F OF ) = 1. Then ε(χ, ψ, µ) = µ(OF ) F × χ−1 · ψ dµ χ ramiﬁed χ unramiﬁed, n(ψ) = 0 Since χ and ψ are locally constant, the integral is actually sum, which turns out to be ﬁnite (this uses the fact that χ is ramiﬁed). For a ∈ F × and b > 0, we have ε(χ, ψ(ax), bµ) = χ(a)\|a\|−1bε(χ, ψ, µ). Theorem (Langlands–Deligne). There exists a unique system of local constants ε(ρ, ψ, µ) for ρ : WF → GLC(V ) such that 37 2 L-functions IV Topics in Number Theory (i) ε is multiplicative in exact sequences, so it is well-deﬁned for virtual representations. (ii) ε(ρ, ψ, bµ) = bdim V ε(ρ, ψ, µ). (iii) If E/F is ﬁnite separable, and ρ is a virtual representation of WF of degree 0 and σ = IndWF WE ρ, then ε(σ, ψ, µ) = ε(ρ, ψ ◦ trE/F , µ). Note that this is independent of the choice of µ and µ, since “dim V = 0”. (iv) If dim ρ = 1, then ε(ρ) is the usual abelian ε(χ). 38 3 -adic representations IV Topics in Number Theory 3 -adic representations In this section, we shall discuss -adic representations of the Galois group, which often naturally arise from geometric situations. At the end of the section, we will relate these to complex representations of the Weil–Langlands group, which will be what enters the Langlands correspondence. Deﬁnition (-adic representation). Let G be a topological group. An -adic representation consists of the following data: – A ﬁnite extension E/Q; – An E-vector space V ; and – A continuous homomorphism ρ : G → GLE(V ) ∼= GLn(E). In this section,we will always take G = ΓF or WF , where F/Qp is a ﬁnite extension with p = . Example. The cyclotomic character χcycl : ΓK → Z× relation ⊆ Q× is deﬁned by the ζ χcycl(γ) = γ(ζ) for all ζ ∈ ¯K with ζ n representation. = 1 and γ ∈ ΓK. This is a one-dimensional -adic Example. Let E/K be an elliptic curve. We deﬁne the Tate module by TE = lim ←− n E[n]( ¯K), VE = TE ⊗Z Q. Then VE is is a 2-dimensional -adic representation of ΓK over Q. Example. More generally, if X/K is any algebraic variety, then V = H i ´et(X ⊗K ¯K, Q) is an -adic representation of ΓK. We will actually focus on the representations of the Weil group WF instead of the full Galois group GF . The reason is that every representation of the Galois group restricts to one of the Weil group, and since the Weil group is dense, no information is lost when doing so. On the other hand, the Weil group can have more representations, and we seek to be slightly more general. Another reason to talk about the Weil group is that local class ﬁeld theory says there is an isomorphism ArtF : W ab F ∼= F ×. So one-dimensional representations of WF are the same as one-dimensional representations of F ×. For example, there is an absolute value map F × → Q×, inducing a representation ω : WF → Q×. Under the Artin map, this sends the geometric Frobenius to 1 q . In fact, ω is the restriction of the cyclotomic character to WF . 39 3 -adic representations IV Topics in Number Theory Recall that we previously deﬁned the tame character. Pick a sequence πn ∈ ¯F by π0 = π and π n+1 = πn. We deﬁned, for any γ ∈ ΓF , t(γ) = γ(πn) πn ∈ lim ←− n µn ( ¯F ) = Z(1). When we restrict to the inertia group, this is a homomorphism, independent of the choice of (πn), which we call the tame character . In fact, this map is ΓF -equivariant, where ΓF acts on IF by conjugation. In general, this still deﬁnes a function ΓF → Z(1), which depends on the choice of πn. Example. Continuing the previous notation, where π is a uniformizer of F , we let Tn be the n-torsion subgroup of ¯F ×/π. Then Tn = ζn, πn/πn n ∼= (Z/nZ)2. The th power map Tn → Tn−1 is surjective, and we can form the inverse limit T , which is then isomorphic to Z2 . This then gives a 2-dimensional -adic representation of ΓF . In terms of the basis (ζn ), (πn), the representation is given by γ → χcycl(γ) 0 . t(γ) 1 Notice that the image of IF is 1 Z 1 0 . In particular, it is inﬁnite. This cannot happen for one-dimensional representations. Perhaps surprisingly, the category of -adic representations of WF over E does not depend on the topology of E, but only the E as an abstract ﬁeld. In particular, if we take E = ¯Q, then after taking care of the slight annoyance that it is inﬁnite over ¯Q, the category of representations over ¯Q does not depend on ! To prove this, we make use of the following understanding of -adic represen- tations. Theorem (Grothendieck’s monodromy theorem). Fix an isomorphism Z(1) ∼= Z. In other words, ﬁx a system (ζn ) such that ζ n = ζn−1 . We then view t as a homomorphism IF → Z via this identiﬁcation. Let ρ : WF → GL(V ) be an -adic representation over E. Then there exists an open subgroup I ⊆ IF and a nilpotent N ∈ EndE V such that for all γ ∈ I , ρ(γ) = exp(t(γ)N ) = ∞ j=0 (t(γ)N )j j! . In particular, ρ(I ) unipotent and abelian. In our previous example, N = 0 0 1 0 . Proof. If ρ(IF ) is ﬁnite, let I = ker ρ ∩ IF and N = 0, and we are done. 40 3 -adic representations IV Topics in Number Theory Otherwise, ﬁrst observe that G is any compact group and ρ : G → GL(V ) is an -adic representation, then V contains a G-invariant lattice, i.e. a ﬁnitelygenerated OE-submodule of maximal rank. To see this, pick any lattice L0 ⊆ V . Then ρ(G)L0 is compact, so generates a lattice which is G-invariant. Thus, pick a basis of an IF -invariant lattice. Then ρ : WF → GLn(E) restricts to a map IF → GLn(OE). We seek to understand this group GLn(OE) better. We deﬁne a ﬁltration on GLn(OE) by Gk = {g ∈ GLn(OE) : g ≡ I mod k}, which is an open subgroup of GLn(OE). Note that for k ≥ 1, there is an isomorphism Gk/Gk+1 → Mn(OE/OE), sending 1 + kg to g. Since the latter is an -group, we know G1 is a pro- group. Also, by deﬁnition, (Gk) ⊆ Gk+1. Since ρ−1(G2) is open, we can pick an open subgroup I ⊆ IF such that ρ(I ) ⊆ G2. Recall that t(IF ) is the maximal pro- quotient of IF , because the tame characters give an isomorphism IF /PF ∼= Z(1). p So ρ\|I : I → G2 factors as t I t(I ) = sZ ν G2 , using the assumption that ρ(IF ) is inﬁnite. Now for r ≥ s, let Tr = ν(r) = T r−s s Nr = log(Tr) = m≥1 ∈ Gr+2−s. For r suﬃciently large, (−1)m−1
(Tr − 1)m m converges -locally, and then Tr = exp Nr. We claim that Nr is nilpotent. To see this, if we enlarge E, we may assume that all the eigenvalues of Nr are in E. For δ ∈ WF and γ ∈ IF , we know So for all γ ∈ I . So t(δγδ−1) = ω(δ)t(γ). ρ(δγδ−1) = ρ(γ)w(σ) ρ(σ)Nrρ(δ−1) = ω(δ)Nr. Choose δ lifting ϕq, w(δ) = q. Then if v is an eigenvector for Nr with eigenvalue λ, then ρ(δ)v is an eigenvector of eigenvalue q−1λ. Since Nr has ﬁnitely many eigenvalues, but we can do this as many times as we like, it must be the case that λ = 0. Then take N = 1 r Nr for r suﬃciently large, and this works. 41 3 -adic representations IV Topics in Number Theory There is a slight unpleasantness in this theorem that we ﬁxed a choice of n Z(1) → roots of unity. To avoid this, we can say there exists an N : v(1) = V ⊗Z V nilpotent such that for all γ ∈ I , we have ρ(γ) = exp(t(γ)N ). Grothendieck’s monodromy theorem motivates the deﬁnition of the Weil– Deligne groups, whose category of representations are isomorphic to the category of -adic representations. It is actually easier to state what the representations of the Weil–Deligne group are. One can then write down a deﬁnition of the Weil–Deligne group as a semi-direct product if they wish. Deﬁnition (Weil–Deligne representation). A Weil–Deligne representation of WF over a ﬁeld E of characteristic 0 is a pair (ρ, N ), where – ρ : WF → GLE(V ) is a ﬁnite-dimensional representation of WF over E with open kernel; and – N ∈ EndE(V ) is nilpotent such that for all γ ∈ WF , we have ρ(γ)N ρ(γ)−1 = ω(γ)N, Note that giving N is the same as giving a unipotent T = exp N , which is the same as giving an algebraic representation of Ga. So a Weil–Deligne representation is a representation of a suitable semi-direct product WF Ga. Weil–Deligne representations form a symmetric monoidal category in the obvious way, with (ρ, N ) ⊗ (ρ, N ) = (ρ ⊗ ρ, N ⊗ 1 + 1 ⊗ N ). There are similarly duals. Theorem. Let E/Q be ﬁnite (and = p). Then there exists an equivalence of (symmetric monoidal) categories -adic representations of WF over E ←→ Weil–Deligne representations of WF over E Note that the left-hand side is pretty topological, while the right-hand side is almost purely algebraic, apart from the requirement that ρ has open kernel. In particular, the topology of E is not used. Proof. We have already ﬁxed an isomorphism Z(1) ∼= Z. We also pick a lift Φ ∈ WF of the geometric Frobenius. In other words, we are picking a splitting WF = Φ IF . The equivalence will take an -adic representation ρ to the Weil–Deligne representation (ρ, N ) on the same vector space such that ρ(Φmγ) = ρ(Φmγ) exp t(γ)N (∗) for all m ∈ Z and γ ∈ IF . 42 3 -adic representations IV Topics in Number Theory To check that this “works”, we ﬁrst look at the right-to-left direction. Suppose we have a Weil–Deligne representation (ρ, N ) on V . We then deﬁne ρ : WF → AutE(V ) by (∗). Since ρ has open kernel, it is continuous. Since t is also continuous, we know ρ is continuous. To see that ρ is a homomorphism, suppose where γ, δ ∈ IF and Φmγ · Φmδ = Φm+nγδ γ = Φ−nγΦn. Then But exp t(γ)N · ρ(Φnδ) = = 1 j! 1 j! j≥0 j≥0 t(γ)jN jρ(Φnδ) t(γ)qnjρ(Φnδ)N j = ρ(Φnδ) exp(qnt(γ)). t(γ) = t(Φ−nγΦn) = ω(Φ−n)t(γ) = qnt(γ). So we know that ρ(Φmγ)ρ(Φnδ) = ρ(Φm+nγδ). Notice that if γ ∈ IF ∩ ker ρ, then ρ(γ) = exp t(γ)N . So N is the nilpotent endomorphism occurring in the Grothendieck theorem. Conversely, given an -adic representation ρ, let N ∈ EndE V be given by the monodromy theorem. We then deﬁne ρ by (∗). Then the same calculation shows that (ρ, N ) is a Weil–Deligne representation, and if I ⊆ IF is the open subgroup occurring in the theorem, then ρ(γ) = exp t(γ)N for all γ ∈ I . So by (∗), we know ρ(I ) = {1}, and so ρ has open kernel. This equivalence depends on two choices — the isomorphism Z(1) ∼= Z and also on the choice of Φ. It is not hard to check that up to natural isomorphisms, the equivalence does not depend on the choices. We can similarly talk about representations over ¯Q, instead of some ﬁnite extension E. Note that if we have a continuous homomorphism ρ : WF → GLn( ¯Q), then there exists a ﬁnite E/Q such that ρ factors through GLn(E). Indeed, ρ(IF ) ⊆ GLn( ¯Q) is compact, since it is a continuous image of compact group. So it is a complete metric space. Moreover, the set of ﬁnite extensions of E/Q is countable (Krasner’s lemma). So by the Baire category theorem, ρ(IF ) is contained in some GLn(E), and of course, ρ(Φ) is contained in some GLn(E). Recalling that a Weil–Deligne representation over E only depends on E as a ∼= ¯Q for any , , we know that ﬁeld, and ¯Q Theorem. Let , = p. Then the category of ¯Q representations of WF is equivalent to the category of ¯Q representations of WF . Conjecturally, -adic representations coming from algebraic geometry have semi-simple Frobenius. This notion is captured by the following proposition/deﬁnition. 43 3 -adic representations IV Topics in Number Theory Proposition. Suppose ρ is an -adic representation corresponding to a Weil– Deligne representation (ρ, N ). Then the following are equivalent: (i) ρ(Φ) is semi-simple (where Φ is a lift of Frobq). (ii) ρ(γ) is semi-simple for all γ ∈ WF \ IF . (iii) ρ is semi-simple. (iv) ρ(Φ) is semi-simple. In this case, we say ρ and (ρ, N ) are F -semisimple (where F refers to Frobenius). ∼= Z IF , and ρ(IF ) is ﬁnite. So that part is always Proof. Recall that WF semisimple, and thus (iii) and (iv) are equivalent. Moreover, since ρ(Φ) = ρ(Φ), we know (i) and (iii) are equivalent. Finally, ρ(Φ) is semi-simple iﬀ ρ(Φn) is semi-simple for all Φ. Then this is equivalent to (ii) since the equivalence before does not depend on the choice of Φ. Example. The Tate module of an elliptic curve over F is not semi-simple, since it has matrix ρ(γ) = ω(γ) 0 . t(γ) 1 However, it is F -semisimple, since ω(γ) 0 ρ(γ . It turns out we can classify all the indecomposable and F -semisimple Weil– Deligne representations. In the case of vector spaces, if a matrix N acts nilpotently on a vector space V , then the Jordan normal form theorem says there is a basis of V in which N takes a particularly nice form, namely the only non-zero entries are the entries of the form (i, i + 1), and the entries are all either 0 or 1. In general, we have the following result: Theorem (Jordan normal form). If V is semi-simple, N ∈ End(V ) is nilpotent with N m+1 = 0, then there exists subobjects P0, . . . , Pm ⊆ V (not unique as subobjects, but unique up to isomorphism), such that N r : Pr → N rPr is an isomorphism, and N r+1Pr = 0, and V = m r=0 Pr ⊕ N Pr ⊕ · · · ⊕ N rPr = m r=0 Pr ⊗Z Z[N ] (N r+1) . For vector spaces, this is just the Jordan normal form for nilpotent matrices. Proof. If we had the desired decomposition, then heuristically, we want to set P0 to be the things killed by N but not in the image of N . Thus, using semisimplicity, we pick P0 to be a splitting ker N = (ker N ∩ im N ) ⊕ P0. Similarly, we can pick P1 by ker N 2 = (ker N + (im N ∩ ker N 2)) ⊕ P1. One then checks that this works. 44 3 -adic representations IV Topics in Number Theory We will apply this when V is a representation of WF → GL(V ) and N is the nilpotent endomorphism of a Weil–Deligne representation. Recall that we had ρ(γ)N ρ(γ)−1 = ω(γ)N, so N is a map V → V ⊗ ω−1, rather than an endomorphism of V . Thankfully, the above result still holds (note that V ⊗ ω−1 is still the same vector space, but with a diﬀerent action of the Weil–Deligne group). Proposition. Let (ρ, N ) be a Weil–Deligne representation. (i) (ρ, N ) is irreducible iﬀ ρ is irreducible and N = 0. (ii) (ρ, N ) is indecomposable and F -semisimple iﬀ (ρ, N ) = (σ, 0) ⊗ sp(n), where σ is an irreducible representation of WF and sp(n) ∼= En is the representation ρ = diag(ωn−1, . . . , ω, 1), Example. If then this is an indecomposable Weil–Deligne representation not of the above form. Proof. (i) is obvious. For (ii), we ﬁrst prove (⇐). If (ρ, N ) = (σ, 0) ⊗ sp(n), then F -semisimplicity is clear, and we have to check that it is indecomposable. Observe that the kernel of N is still a representation of WF . Writing V N =0 for the kernel of N in V , we note that V N =0 = σ ⊗ ωn−1, which is irreducible. Suppose that (ρ, N ) = U1 ⊕ U2. Then for each i, we must have U N =0 U N =0 1 = 0. Then this forces U1 = 0. So we are done. i = 0 or V N =0. We may wlog assume Conversely, if (ρ, N, V ) is F -semisimple and indecomposable, then V is a representation of WF which is semi-simple and N : V → V ⊗ ω−1. By Jordan normal form, we must have rU with N r+1 = 0, and U is irreducible. So V = (σ, 0) ⊗ sp(r + 1). Given this classiﬁcation result, when working over complex representations, the representation theory of SU(2) lets us capture the N bit of F -semisimple Weil–Deligne representation via the following group: 45 3 -adic representations IV Topics in Number Theory Deﬁnition (Weil–Langlands group). We deﬁne the (Weil–)Langlands group to be LF = WF × SU(2). A representation of LF is a continuous action on a ﬁnite-dimensional vector space (thus, the restriction to WF has open kernel). Theorem. There exists a bijection between F -semisimple Weil–Deligne representations over C and semi-simple representations of LF , compatible with tensor products, duals, dimension, etc. In this correspondence: – The representations ρ of LF that factor through WF correspond to the Weil–Deligne representations (ρ, 0). – More generally, simple LF representations σ ⊗ (Symn−1 C2) correspond to the Weil–Deligne representation (σ ⊗ ω(−1+n)/2, 0) ⊗ sp(n). If one sits down and checks the theorem, then one sees that the twist in the second part is required to ensure compatibility with tensor products. Of course, the (F -semisimple) Weil–Deligne representations over C are in bijection those over ¯Q, using an isomorphism ¯Q ∼= C. 46 4 The Langlands correspondence IV Topics in Number Theory 4 The Langlands correspondence Local class ﬁeld theory says we have an isomorphism W ab F ∼= F ×. If we want to state th
is in terms of the full Weil group, we can talk about the one-dimensional representations of WF , and write local class ﬁeld theory as a correspondence characters of GL1(F ) ←→ 1-dimensional representations of WF The Langlands correspondence aims to understand the representations of GLn(F ), and it turns out this corresponds to n-dimensional representations of LF . That is, if we put enough adjectives in front of these words. 4.1 Representations of groups The adjectives we need are fairly general. The group GLn(F ) contains a proﬁnite open subgroup GLn(OF ). The general theory applies to any topological group with a proﬁnite open subgroup K, with G/K countable. Deﬁnition (Smooth representation). A smooth representation of G is a continuous representation of G over C, where C is given the discrete topology. That is, it is a pair (π, V ) where V is a complex vector space and π : G → GLC(V ) a homomorphism such that for every v ∈ V , the stabilizer of v in G is open. Note that we can replace C with any ﬁeld, but we like C. Typically, V is an inﬁnite-dimensional vector space. To retain some sanity, we often desire the following property: Deﬁnition (Admissible representation). We say (π, V ) is admissible if for every open compact subgroup K ⊆ G the ﬁxed set V K = {v ∈ V : π(g)v = v ∀g ∈ K} is ﬁnite-dimensional. Example. Take G = GL2(F ). Then P1(F ) has a right action of G by linear transformations. In fact, we can write P1(F ) as P1(F ) = ∗ 0 ∗ ∗ \ G. Let V be the space of all locally constant functions f : P1(F ) → C. There are lots of such functions, because P1(F ) is totally disconnected. However, since P1(F ) is compact, each such function can only take ﬁnitely many values. We let π(g)f = (x → f (xg)). It is not diﬃcult to see that this is an inﬁnite-dimensional admissible representation. Of course, any ﬁnite-dimensional representation is an example, but GLn(F ) does not have very interesting ﬁnite-dimensional representations. 47 4 The Langlands correspondence IV Topics in Number Theory Proposition. Let G = GLn(F ). dim V < ∞, then If (π, V ) is a smooth representation with π = σ ◦ det for some σ : F × → GLC(V ). So these are pretty boring. Proof. If V = d i=1 Cei, then d ker π = (stabilizers of ei) i=1 is open. It is also a normal subgroup, so ker π ⊇ Km = {g ∈ GLn(O) : g ≡ I mod m} for some m, where is a uniformizer of F . In particular, ker π contains Eij(x) for some i = j and x, which is the matrix that is the identity except at entry (i, j), where it is x. But since ker π is normal, conjugation by diagonal matrices shows that it contains all Eij(x) for all x ∈ F and i = j. For any ﬁeld, these matrices generate SLn(F ). So we are done. So the interesting representations are all inﬁnite-dimensional. Fortunately, a lot of things true for ﬁnite-dimensional representations also hold for these. For example, Lemma (Schur’s lemma). Let (π, V ) be an irreducible representation. Then every endomorphism of V commuting with π is a scalar. In particular, there exists ωπ : Z(G) → C× such that π(zg) = ωπ(z)π(g) for all z ∈ Z(G) and g ∈ G. This is called the central character . At this point, we are already well-equipped to state a high-level description of the local Langlands correspondence. Theorem (Harris–Taylor, Henniart). There is a bijection irreducible, admissible representations of GLn(F ) ←→ semi-simple n-dimensional representations of LF . In the next section, we will introduce the Hecke algebra, which allows us to capture these scary inﬁnite dimensional representations of GLn(F ) in terms of something ﬁnite-dimensional. Afterwards, we are going to state the Langlands classiﬁcation of irreducible admissible representations of GLn(F ). We can then state a detailed version of the local Langlands correspondence in terms of this classiﬁcation. 48 4 The Langlands correspondence IV Topics in Number Theory 4.2 Hecke algebras Let G, K be as before. Notation. We write C∞ f : G → C of compact support. c (G) for the vector space of locally constant functions Deﬁnition (Hecke algebra). The Hecke algebra is deﬁned to be H(G, K) = {ϕ ∈ C∞ c (G) : ϕ(kgk) = ϕ(g) for all k, k ∈ K}. This is spanned by the characteristic functions of double cosets KgK. This algebra comes with a product called the convolution product. To deﬁne c (G) → C, written this, we need the Haar measure on G. This is a functional C∞ f → G f (g) dµ(g), that is invariant under left translation, i.e. for all h ∈ G,we have f (hg) dµ(g) = f (g) dµ(g). To construct the Haar measure, we take µ(1K) = 1. Then if K ⊆ K is an open subgroup, then it is of ﬁnite index, and since we want µ(1xK) = µ(1K), we must have µ(1K) = 1 (K : K ) . We then set µ(1xK) = µ(1K) for any x ∈ G, and since these form a basis of the topology, this deﬁnes µ. Deﬁnition (Convolution product). The convolution product on H(G, K) is (ϕ ∗ ϕ)(g) = G ϕ(x)ϕ(x−1g) dµ(x). Observe that this integral is actually a ﬁnite sum. It is an exercise to check that this is a C-algebra with unit eK = 1 µ(K) 1K, Now if (π, V ) is a smooth representation, then for all v ∈ V and ϕ ∈ H(G, K), consider the expression π(ϕ)v = G ϕ(g)π(g)v dµ(g). Note that since the stabilizer of v is open, the integral is actually a ﬁnite sum, so we can make sense of it. One then sees that π(ϕ)π(ϕ) = π(ϕ ∗ ϕ). 49 4 The Langlands correspondence IV Topics in Number Theory This would imply V is a H(G, K)-module, if the unit acted appropriately. It doesn’t, however, since in fact π(ϕ) always maps into V K. Indeed, if k ∈ K, then π(k)π(ϕ)v = ϕ(g)π(kg)v dµ(g) = G ϕ(g)π(g) dµ(g) = π(ϕ)v, using that ϕ(g) = ϕ(k−1g) and dµ(g) = dµ(k−1g). So our best hope is that V K is an H(G, K)-module, and one easily checks that π(eK) indeed acts as the identity. We also have a canonical projection π(eK) : V V K. In good situations, this Hecke module determines V . Proposition. There is a bijection between isomorphism classes of irreducible admissible (π, V ) with V K = 0 and isomorphism classes of simple ﬁnite-dimensional H(G, K)-modules, which sends (π, V ) to V K with the action we described. If we replace K by a smaller subgroup K ⊆ K, then we have an inclusion H(G, K) → H(G, K ), which does not take eK to eK. We can then form the union of all of these, and let H(G) = lim −→ K H(G, K) which is an algebra without unit. Heuristically, the unit should be the delta function concentrated at the identity, but that is not a function. This H(G) acts on any smooth representation, and we get an equivalence of categories smooth G-representations ←→ non-degenerate H(G)-modules . The non-degeneracy condition is V = H(G)V . Note that if ϕ ∈ H(G) and (π, V ) is admissible, then the rank π(ϕ) < ∞, using that V K is ﬁnite-dimensional. So the trace is well-deﬁned. The character of (π, V ) is then the map In this sense, admissible representations have traces. ϕ → tr π(ϕ) 4.3 The Langlands classiﬁcation Recall that the group algebra C[G] is an important tool in the representation theory of ﬁnite groups. This decomposes as a direct sum over all irreducible representations C[G] = πdim(π). π The same result is true for compact groups, if we replace C[G] by L2(G). We get a decomposition L2(G) = πdim(π), ˆ π 50 4 The Langlands correspondence IV Topics in Number Theory where L2 is deﬁned with respect to the Haar measure, and the sum is over all (ﬁnite dimensional) irreducible representations of G. The hat on the direct sum says it is a Hilbert space direct sum, which is the completion of the vector space direct sum. This result is known as the Peter–Weyl theorem. For example, L2(R/Z) = ˆ C · e2πiny. n∈Z However, if G is non-compact, then this is no longer true. Sometimes, we can salvage this a bit by replacing the discrete direct sum with a continuous version. For example, the characters of R are those of the form x → e2πixy, which are not L2 functions. But we can write any function in L2(R) as x → y ϕ(y)e2πixy dy. So in a sense, L2(R) is the “continuous direct sum” of irreducible representations. In general, L2(G) decomposes as a sum of irreducible representations, and contains both a discrete sum and a continuous part. However, there are irreducible representations that don’t appear in L2(G), discretely or continuously. These are known as the complementary series representations. This happens, for example, for G = SL2(R) (Bargmann 1947). We now focus on the case G = GLn(F ), or any reductive F -group (it doesn’t hurt to go for generality if we are not proving anything anyway). It turns out in this case, we can describe the representations that appear in L2(G) pretty explicitly. These are characterized by the matrix coeﬃcients. If π : G → GLn(C) is a ﬁnite-dimensional representation, then the matrix coeﬃcients π(g)ij can be written as π(g)ij = δi(π(g)ej), where ej ∈ Cn is the jth basis vector and δi ∈ (Cn)∗ is the ith dual basis vector. More generally, if π : G → GL(V ) is a ﬁnite-dimensional representation, and v ∈ V , ∈ V ∗, then we can think of πv,(g) = (π(g)v) as a matrix element of π(g), and this deﬁnes a function πv, : G → C. In the case of the G we care about, our representations are fancy inﬁnitedimensional representations, and we need a fancy version of the dual known as the contragredient. Deﬁnition (Contragredient). Let (π, V ) be a smooth representation. We deﬁne V ∗ = HomC(V, C), and the representation (π∗, V ∗) of G is deﬁned by π∗(g) = (v → (π(g−1)v)). We then deﬁne the contragredient (˜π, ˜v) to be the subrepresentation ˜V = { ∈ V ∗ with open stabilizer}. 51 4 The Langlands correspondence IV Topics in Number Theory This contragredient is quite pleasant. Recall that We then have V = V K. K ˜V = (V ∗)K. Using the projection π(eK) : V → V K, we can identify (V ∗)K = (V K)∗. So in particular, if π is admissible, then so is ˜π, and we have a canonical isomorphism V → ˜˜V. Deﬁnition (Matrix coeﬃcient). Let (π, V ) be a smooth representation, and v ∈ V , ∈ ˜V . The matrix coeﬃcient πv, is deﬁned by πv,(g) = (π(g)v). This is a locally constant function G → C. We can now mak
e the following deﬁnition: Deﬁnition (Square integrable representation). Let (π, V ) be an irreducible smooth representation of G. We say it is square integrable if ωπ is unitary and for all (v, ). \|πv,\| ∈ L2(G/Z) Note that the fact that ωπ is unitary implies \|πv,\| is indeed a function on L2(G/Z). In general, it is unlikely that πv, is in L2(G). If Z is ﬁnite, then ωπ is automatically unitary and we don’t have to worry about quotienting about the center. Moreover, π is square integrable iﬀ πv, ∈ L2(G). In this case, if we pick ∈ ˜V non-zero, then v → πv, gives an embedding of V into L2(G). In general, we have V ⊆ L2(G, ωπ) = {f : G → C : f (zg) = ωπ(z)f (z), \|f \| ∈ L2(G/Z)}, A slight weakening of square integrability is the following weird deﬁnition: Deﬁnition (Tempered representation). Let (π, V ) be irreducible, ωπ unitary. We say it is tempered if for all (v, ) and ε > 0, we have \|πv,\| ∈ L2+ε(G/Z). The reason for this deﬁnition is that π is tempered iﬀ it occurs in L2(G), not necessarily discretely. Weakening in another direction gives the following deﬁnition: Deﬁnition (Essentially square integrable). Let (π, V ) be irreducible. Then (π, V ) is essentially square integrable (or essentially tempered ) if is square integrable (or tempered) for some character χ : F × → C. π ⊗ (χ ◦ det) 52 4 The Langlands correspondence IV Topics in Number Theory Note that while these deﬁnitions seem very analytic, there are in fact purely algebraic interpretations of these deﬁnitions, using Jacquet modules. A ﬁnal category of representations is the following: Deﬁnition (Supercuspidal representation). We say π is supercuspidal if for all (v, ), the support of πv, is compact mod Z. These are important because they are building blocks of all irreducible representations of GLn(F ), in a sense we will make precise. The key notion is that of parabolic induction, which takes a list of represen- tations σi of GLni(F ) to a representation of GLN (F ), where N = ni. We ﬁrst consider a simpler case, where we have an n-tuple χ = (χ1, . . . , χn) : (F ×)n → C× of characters. The group G = GLn(F ) containing the Borel subgroup B of upper-triangular matrices. Then B is the semi-direct product T N , where T ∼= (F ×)n consists of the diagonal matrices and N the unipotent ones. We can then view χ as a character χ : B → B/N = T → C×. We then induce this up to a representation of G. Here the deﬁnition of an induced representation is not the usual one, but has a twist. Deﬁnition (Induced representation). Let χ : B → C be a character. We deﬁne the induced representation IndG B(χ) to be the space of locally constant functions f : g → C such that for all b ∈ B and g ∈ G, where G acts by f (bg) = χ(b)δB(b)1/2f (g) π(g)f : x → f (xg). The function δB(b)1/2 is deﬁned by δB(b) = \| det adB(b)\|. More explicitly, if the diagonal entries of b ∈ B are x1, . . . , xn, then δB(b) = n i=1 \|xi\|n+1−2i = \|x1\|n−1\|x2\|n−3 · · · \|xn\|−n+1 This is a smooth representation since B\G is compact. In fact, it is admissible and of ﬁnite length. When this is irreducible, it is said to be a principle series representation of G. Example. Recall that P1(F ) = B\GL2(F ). In this terminology, C∞(P1(F )) = IndG B(δ−1/2 B ). This is not irreducible, since it contains the constant functions, but quotienting by these does give an irreducible representation. This is called the Steinberg representation. In general, we start with a parabolic subgroup P ⊆ GLn(F ) = G, i.e. one conjugate to block upper diagonal matrices with a partition n = n1 + · · · + nr. This then decomposes into M N , where       In1 · · · . . . ∗ ... Inr       . M ∼= i GLni(F ), N = 53 4 The Langlands correspondence IV Topics in Number Theory This is an example of a Levi decomposition, and M is a Levi subgroup. To perform parabolic induction, we let (σ, U ) be a smooth representation of M , e.g. σ1 ⊗ · · · ⊗ σr, where each σi is a representation of GLni(F ). This then deﬁnes a representation of P via P → P/N = M , and we deﬁne IndG P (σ) to be the space of all locally constant functions f : G → U such that f (pg) = δP (p)σ(p)f (g) for all p ∈ P, g ∈ G and δP is again deﬁned by δP (p) = \| det adP (p)\|. This is again a smooth representation. Proposition. (i) σ is admissible implies π = IndG P σ is admissible. (ii) σ is unitary implies π is unitary. (iii) IndG P (˜σ) = ˜π. (ii) and (iii) are the reasons for the factor δ1/2 P . Example. Observe C∞(P(F )) = Ind(δ1/2 B ) = {f : G → C : f (bg) = δB(b)f (g)}. There is a linear form to C given by integrating f over GL2(O) (we can’t use F since GL2(F ) is not compact). In fact, this map is G-invariant, and not just GL2(O)-invariant. This is dual to the constant subspace of C∞(P1(F )). The rough statement of the classiﬁcation theorem is that every irreducible admissible representation of G = GLn(F ), is a subquotient of an induced representation IndG P σ for some supercuspidal representation of a Levi subgroup P = GLn1 × · · · × GLnr (F ). This holds for any reductive G/F . For GLn(F ), we can be more precise. This is called the Langlands classiﬁca- tion. We ﬁrst classify all the essentially square integrable representations: Theorem. Let n = mr with m, r ≥ 1. Let σ be any supercuspidal representation of GLm(F ). Let σ(x) = σ ⊗ \|detm\|x. Write ∆ = (σ, σ(1), . . . , σ(r − 1)), a representation of GLm(F ) × · · · × GLm(F ). Then IndG P (∆) has a unique irreducible subquotient Q(∆), which is essentially square integrable. Moreover, Q(∆) is square integrable iﬀ the central character is unitary, iﬀ 2 ) is square-integrable, and every essentially square integrable π is a Q(∆) σ( r−1 for a unique ∆. Example. Take n = 2 = r, σ = \| · \|−1/2. Take P = B = ∗ 0 ∗ ∗ 54 4 The Langlands correspondence IV Topics in Number Theory Then IndG B(\| · \|−1/2, \| · \|−1/2) = C∞(B\G) ⊇ C, where C is the constants. Then C∞(B\G) is not supercuspidal, but the quotient is the Steinberg representation, which is square integrable. Thus, every two-dimensional essentially square integrable representation which is not supercuspidal is a twist of the Steinberg representation by χ ◦ det. We can next classify tempered representations. Theorem. The tempered irreducible admissible representations of GLn(F ) are precisely the representations IndG P σ, where σ is irreducible square integrable. In particular, IndG P σ are always irreducible when σ is square integrable. Example. For GL2, we seek a π which is tempered but not square integrable. This must be of the form π = IndG B(χ1, χ2), where \|χ1\| = \|χ2\| = 1. If we want it to be essentially tempered, then we only need \|χ1\| = \|χ2\|. Finally, we classify all irreducible (admissible) representations. Theorem. Let n = n1 + · · · + nr be a partition, and σi tempered representation of GLni(F ). Let ti ∈ R with t1 > · · · > tr. Then IndG P (σ1(t1), . . . , σr(tr)) has a unique irreducible quotient Langlands quotient, and every π is (uniquely) of this form. Example. For GL2, the remaining (i.e. not essentially tempered) representations are the irreducible subquotients of IndG B(χ1, χ2), where \|χi\| = \| · \|ti F , Note that the one-dimensional representations must occur in this set, because we haven’t encountered any yet. t1 > t2. For example, if we take χ1 = \| · \|1/2 and χ2 = \| · \|−1/2, then IndG B(χ1, χ2) = C∞(B\G), which has the trivial representation as its irreducible quotient. 4.4 Local Langlands correspondence Theorem (Harris–Taylor, Henniart). There is a bijection irreducible, admissible representations of GLn(F ) ←→ semi-simple n-dimensional representations of LF . Moreover, – For n = 1, this is the same as local class ﬁeld theory. – Under local class ﬁeld theory, this corresponds between ωπ and det σ. 55 4 The Langlands correspondence IV Topics in Number Theory – The supercuspidals correspond to the irreducible representations of WF itself. – If a supercuspidal π0 corresponds to the representation σ0 of WF , then the 2 )) essentially square integrable representation π = Q(π0(− r−1 corresponds to σ = σ0 ⊗ Symr−1 C2. 2 ), . . . , π0( r−1 – If πi correspond to σi, where σi are irreducible and unitary, then the P (π1 ⊗ · · · ⊗ πr) corresponds to σ1 ⊕ · · · ⊕ σr. tempered representation IndG – For general representations, if π is the Langlands quotient of Ind(π1(t1), . . . , πr(tr)) with each πi tempered, and πi corresponds to unitary representations σi of LF , then π corresponds to σi ⊗ \|Art−1 F \|ti F . The hard part of the theorem is the correspondence between the supercuspidal representations and irreducible representations of WF . This correspondence is characterized by ε-factors of pairs. Recall that for an irreducible representation of WF , we had an ε factor ε(σ, µF , ψ). If we have two representations, then we can just take the tensor product ε(σ1 ⊗ σ2, µF , ψ). It turns out for supercuspidals, we can also introduce ε-factors ε(π, µF , ψ). There are also ε factors for pairs, ε(π1, π2, µF , ψ). Then the correspondence is such that if πi correspond to σi, then ε(σ1 ⊗ σ2, µF , ψ) = ε(π1, π2, µF , ψ). When n = 1, we get local class ﬁeld theory. Recall that we actually have a homomorphic correspondence between characters of F × and characters of WF , and the correspondence is uniquely determined by (i) The behaviour on unramiﬁed characters, which is saying that the Artin map sends uniformizers to geometric Frobenii (ii) The base change property: the restriction map W ab F → W ab F correspond to the norm map of ﬁelds If we want to extend this to the local Langlands correspondence, the corresponding picture will include (i) Multiplication: taking multiplications of GLn and GLm to representations of GLmn (ii) Base change: sending representations of GLn(F ) to representations of GLn(F ) for a ﬁnite extension F /F These thing exist by virtue of local Langlands correspondence (much earlier, base change for cyclic extensions was constructed by Arthur–Clozel). Proposition. Let σ : WF → GLn(C) be an irreducible representation. Then the following are equivalent: (i) For so
me g ∈ WF \ IF , σ(g) has an eigenvalue of absolute value 1. 56 4 The Langlands correspondence IV Topics in Number Theory (ii) im σ is relatively compact, i.e. has compact closure, i.e. is bounded. (iii) σ is unitary. Proof. The only non-trivial part is (i) ⇒ (ii). We know im σ = σ(Φ), σ(IF ) = H, where Φ is some lift of the Frobenius and H is a ﬁnite group. Moreover, IF is normal in WF . So for some n ≥ 1, σ(Φn) commutes with H. Thus, replacing g and Φn with suitable non-zero powers, we can assume σ(g) = σ(Φn)h for some h ∈ H. Since H is ﬁnite, and σ(Φn) commutes with h, we may in fact assume σ(g) = σ(Φ)n. So we know σ(Φ) has eigenvalue with absolute value 1. Let V1 ⊆ V = Cn be a sum of eigenspaces for σ(Φ)n with all eigenvalues having absolute value 1. Since σ(Φn) is central, we know V1 is invariant, and hence V is irreducible. So V1 = V . So all eigenvalues of σ(Φ) have eigenvalue 1. Since V is irreducible, we know it is F -semisimple. So σ(Φ) is semisimple. So σ(Φ) is bounded. So im σ is bounded. 57 5 Modular forms and representation theory IV Topics in Number Theory 5 Modular forms and representation theory Recall that a modular form is a holomorphic function f : H → C such that f (z) = j(γ, z)−kf (γ(z)) for all γ in some congruence subgroup of SL2(Z), where γ = a b d c , γ(z) = az + b cz + d , j(γ, z) = cz + d. Let Mk be the set of all such f . Consider the group GL2(Q)+ = {g ∈ GL2(Q) : det g > 0}. This acts on Mk on the left by g : f → j1(g−1, z)−kf (g−1(z)), j1(g, z) = \| det g\|−1/2j(g, z). The factor of \| det g\|−1/2 makes the diagonal diag(Q× >0) act trivially. Note that later we will consider some g with negative determinant, so we put the absolute value sign in. For any f ∈ Mk, the stabilizer contains some Γ(N ), which we can think of as some continuity condition. To make this precise, we can form the completion of GL2(Q)+ with respect to {Γ(N ) : N ≥ 1}, and the action extends to a representation π of this completion. In fact, this completion is G = {g ∈ GL2(A∞ Q ) : det(g) ∈ Q× >0}. This is a closed subgroup of G = GL2(A∞ Q ), and in fact G = G · ˆZ 0 1 0 . In fact G is a semidirect product of the groups. The group G seems quite nasty, since the determinant condition is rather unnatural. It would be nice to get a representation of G itself, and the easy way to do so is by induction. What is this? By deﬁnition, it is IndG G(Mk) = {ϕ : G → Mk : ∀h ∈ G, ϕ(hg) = π(h)ϕ(g)}. Equivalently, this consists of functions F : H × G → C such that for all γ ∈ GL2(Q)+, we have j1(γ, z)−kF (γ(z), γg) = F (z, g), and for every g ∈ G, there is some open compact K ⊆ G such that F (z, g) = F (z, gh) for all h ∈ K, and that F is holomorphic in z (and at the cusps). To get rid of the plus, we can just replace GL2(Q)+, H with GL2(Q), C \ R = H±. These objects are called adelic modular forms. 58 5 Modular forms and representation theory IV Topics in Number Theory If F is an adelic modular form, and we ﬁx g, then the function f (z) = F (z, g) is a usual modular form. Conversely, if F invariant under ker(G2(ˆZ) → GL2(Z/N Z)), then F corresponds to a tuple of Γ(N )-modular forms indexed by (Z/N Z)×. This has an action of G = p GL2(Qp) (which is the restricted product with respect to GL2(Zp)). The adelic modular forms contain the cusp forms, and any f ∈ Mk generates a subrepresentation of the space of adelic forms. Theorem. (i) The space Vf of adelic cusp forms generated by f ∈ Sk(Γ1(N )) is irreducible iﬀ f is a Tp eigenvector for all p n. (ii) This gives a bijection between irreducible G-invariant spaces of adelic cusp forms and Atkin–Lehner newforms. Note that it is important to stick to cusp forms, where there is an inner product, because if we look at the space of adelic modular forms, it is not completely decomposable. Now suppose (π, V ) is an irreducible admissible representation of GL2(A∞ Q ) = Gp = GL2(Qp), and take a maximal compact subgroups K 0 p = GL2(Zp) ⊆ GL2(Qp). Then general facts about irreducible representations of products imply irreducibility (and admissibility) is equivalent to the existence of irreducible admissible representations (πp, Vp) of Gp for all p such that (i) For almost all p, dim V K0 p ≥ 1 (for Gp = GLn(Qp), this implies the p dimension is 1). Fix some non-zero f 0 p ∈ V K0 p p . (ii) We have π = ⊗ pπp, the restricted tensor product, which is generated by for almost all p. To be precise, p vp with vp = f 0 p ⊗ pπp = lim −→ ﬁnite S p∈S πp. The use of vp is to identify smaller tensor products with larger tensor products. Note that (i) is equivalent to the assertion that (πp, Vp) is an irreducible (χ1, χ2) where χi are unramiﬁed characters. principal series representation IndGp Bp These unramiﬁed characters are determined by χp(p) = αp,i. If f = anqn ∈ Sk(Γ1(N )) is a normalized eigenform with character ω : (Z/N Z)× → C×, and if f corresponds to π = πp, then for every p N , we have πp = IndGp (χ1, χ2) is an unramiﬁed principal series, and Bp ap = p(k−1)/2(αp,1 + αp,2) ω(p)−1 = αp,1αp,2. We can now translate diﬃcult theorems about modular forms into representation theory. 59 5 Modular forms and representation theory IV Topics in Number Theory Example. The Ramanujan conjecture (proved by Deligne) says \|ap\| ≤ 2p(k−1)/2. If we look at the formulae above, since ω(p)−1 is a root of unity, this is equivalent to the statement that \|αp,i\| = 1. This is true iﬀ πp is tempered. We said above that if dim V K0 p p ≥ 1, then in fact it is equal to 1. There is a generalization of that. Theorem (Local Atkin–Lehner theorem). If (π, V ) is an irreducible representation of GL2(F ), where F/Qp and dim V = ∞, then there exists a unique nπ > 0 such that V Kn = 0 n < nπ one-dimensional n = nπ , Kn = g ≡ ∗ 0 ∗ 1 mod n . Taking the product of these invariant vectors for n = nπ over all p gives Atkin– Lehner newform. What about the other primes? i.e. the primes at inﬁnity? We have a map f : H± × GL2(A∞ Q ) → C. Writing H = SL2(R)/SO(2), which has an action of Γ, we can convert this to an action of SO(2) on Γ\SL2(R). Consider the function Φf : GL2(R) × GL2(A∞ Q ) = GL2(AQ) → C given by Φf (h∞, h∞) = j1(h∞, i)−kf (h∞(i), h∞). Then this is invariant under GL2(Q)+, i.e. Now if we take then Φf (γh∞, γh∞) = Φ(h∞, h∞). kθ = cos θ − sin θ sin θ cos θ ∈ SO(2), Φf (h∞kθ, h∞) = eikθΦf (h∞, h∞). So we get invariance under γ, but we need to change what SO(2) does. In other words, Φf is now a function Φf : GL2(Q)\G2(AQ) → C satisfying various properties: – It generates a ﬁnite-dimensional representation of SO(2) – It is invariant under an open subset of GL2(A∞ Q ) – It satisﬁes growth condition and cuspidality, etc. 60 5 Modular forms and representation theory IV Topics in Number Theory By the Cauchy–Riemann equations, the holomorphicity condition of f says Φf satisﬁes some diﬀerential equation. In particular, that says Φf is an eigenfunction for the Casimir in the universal enveloping algebra of sl2. These conditions together deﬁne the space of automorphic forms. Example. Take the non-holomorphic Eisenstein series E(z, s) = (c,d)=(0,0) 1 \|cz + d\|2s . This is a real analytic function on Γ\H → C. Using the above process, we get an automorphic form on GL2(Q)\GL2(AQ) with k = 0. So it actually invariant under SO(2). It satisﬁes ∆E = s(1 − s)E. There exist automorphic cusp forms which are invariant under SO(2), known as Maass forms. They also satisfy a diﬀerential equation with a Laplacian eigenvalue λ. A famous conjecture of Selberg says λ ≥ 1 4 . What this is equivalent to is that the representation of GL2(R) they generate is tempered. In terms of representation theory, this looks quite similar to Ramanujan’s conjecture! 61 Index Index ∞(G), 49 K, 13 C c C 0 F -semisimple, 44 F ur, 5 IF , 5 IS, 18 JK, 12 K ab, 4 L(χ, s), 30 PF , 5 ArtF , 6 Frobq, 5 ΓK, 4 Λ(χ, s), 30 Σ∞ K , 11 ΣK,∞, 11 ¯K, 4 -adic representation, 39 Z(1)(¯k), 6 AK, 12 A× K, 12 H(G), 50 H(G, K), 49 a(∞), 17 µn(¯k), 5 πv,, 52 kF , 5 v \| ∞, 11 v ∞, 11 adele, 12 adelic modular form, 58 admissible representation, 47 algebraic Hecke character, 24 algebraic homomorphism, 23 Archimedean local ﬁeld, 4 arithmetic Frobenius, 5 Artin map global, 13 local, 6 Artin reciprocity law, 14 automorphic forms, 61 Brauer group, 9 central character, 48 CM ﬁeld, 27 IV Topics in Number Theory complementary series, 51 conductor, 18 connected Serre group, 29 content homomorphism, 16 contragredient, 51 convolution product, 49 cyclotomic character, 39 decomposition group, 13 essentially square integrable, 52 essentially tempered, 52 existence theorem, 7 ﬁnite place, 11 fundamental mistake of class ﬁeld theory, 18 generalized ideal class group, 18 geometric Frobenius, 5 global ε-factor, 30 global Artin map, 13 global ﬁeld, 11 Gr¨oßencharakter, 20 Grothendieck monodromy theorem, 40 Hecke L-function, 29 Hecke algebra, 49 Hecke character, 20 idele, 12 idele class group, 12 induced representation, 53 inertia group, 5 inﬁnite place, 11 inﬁnite type, 24 Kronecker–Weber theorem, 17 Krull topology, 4 Langlands correspondence, 48 Langlands group, 46 Langlands quotient, 55 Levi decomposition, 54 Levi subgroup, 54 local ε-factors, 31 local Artin map, 6 62 Index IV Topics in Number Theory local class ﬁeld theory, 6 local ﬁeld, 4 local Langlands correspondence, 55 ray class group, 18 relative Weil group, 9 residue ﬁeld, 5 matrix coeﬃcient, 52 maximal tamely ramiﬁed extension, 6 maximal unramiﬁed extension, 5 modulus, 17 non-Archimedean local ﬁeld, 4 parabolic induction, 54 parabolic subgroup, 53 Peter–Weyl theorem, 51 place, 11 principle series representation, 53 product formula, 11 proﬁnite group, 4 quasi-characters, 20 ramiﬁcation real place, 15 ray class character, 21 ray class ﬁeld, 18 Schur’s lemma, 48 Serre type, 27 smooth representation, 47 square integrable representation, 52 Steinberg representation, 53 supercuspidal representation, 53 tame character, 5, 40 Tamiyama group, 29 Tate module, 6, 39 tempered representation, 52 uniformizer, 4 unramiﬁed character, 21 valuation ring, 4 Weil group, 8 Weil–Delig
ne representation, 42 Weil–Langlands group, 46 wild inertia group, 5 63
D n n m! C m ; 1! m 0 n; and use this to show by induction on n that n m! D nŠ mŠ.n ; 0 m n: m/Š (b) Show that n 0 m X D 1/m . n m! D 0 and n 0 m X D n m! D 2n: (c) Show that y/n .x C D (This is the binomial theorem.) n 0 m X D n m! m: xmyn 20. Use induction to ﬁnd an nth antiderivative of log x, the natural logarithm of x. 21. Let f1.x1/ x1. For n g1.x1/ 2, let D D fn.x1; x2; : : : ; xn/ fn 1.x1; x2; : : : ; xn fn 1/ 1.x1; x2; : : : ; xn C 1/ 2xn 2n C 2xn 2n j j and gn.x1; x2; : : : ; xn/ j Find explicit formulas for fn.x1; x2; : : : ; xn/ and gn.x1; x2; : : : ; xn/. gn 1.x1; x2; : : : ; xn gn 1/ 1.x1; x2; : : : ; xn C 1/ 2xn 2n 2xn 2n j : D D 22. Prove by induction that sin x C sin 3x C C sin.2n 1/x D 1 cos 2nx 2 sin x ; n 1: HINT: You will need trigonometric identities that you can derive from the identities cos.A cos.A B/ B/ C D D cos A cos B cos A cos B C sin A sin B; sin A sin B: Take these two identities as given: 23. Suppose that a1 permutation of f a2 1; 2; : : : ; n g an and b1 , and deﬁne Q.`1; `2; : : : ; `n/ Show that Section 1.3 TheReal Line 19 b2 bn. Let f `1; `2; : : : `n be a g n D 1 Xi D .ai b`i /2: Q.`1; `2; : : : ; `n/ Q.1; 2; : : : ; n/: 1.3 THE REAL LINE One of our objectives is to develop rigorously the concepts of limit, continuity, differentiability, and integrability, which you have seen in calculus. To do this requires a better understanding of the real numbers than is provided in calculus. The purpose of this section is to develop this understanding. Since the utility of the concepts introduced here will not become apparent until we are well into the study of limits and continuity, you should reserve judgment on their value until they are applied. As this occurs, you should reread the applicable parts of this section. This applies especially to the concept of an open covering and to the Heine–Borel and Bolzano–Weierstrass theorems, which will seem mysterious at ﬁrst. We assume that you are familiar with the geometric interpretation of the real numbers as points on a line. We will not prove that this interpretation is legitimate, for two reasons: (1) the proof requires an excursion into the foundations of Euclidean geometry, which is not the purpose of this book; (2) although we will use geometric terminology and intuition in discussing the reals, we will base all proofs on properties (A)–(I) (Section 1.1) and their consequences, not on geometric arguments. Henceforth, we will use the terms real number system and real line synonymously and denote both by the symbol R; also, we will often refer to a real number as a point (on the real line). Some Set Theory In this section we are interested in sets of points on the real line; however, we will consider other kinds of sets in subsequent sections. The following deﬁnition applies to arbitrary sets, with the understanding that the members of all sets under consideration in any given context come from a speciﬁc collection of elements, called the universal set. In this section the universal set is the real numbers. Deﬁnition 1.3.1 Let S and T be sets. (a) S contains T , and we write S T or T this case, T is a subset of S . S , if every member of T is also in S . In T is the set of elements that are in S but not in T . (b) S (c) S equals T , and we write S and only if S and T have the same members. D T , if S contains T and T contains S ; thus, S T if D 20 Chapter 1 TheRealNumbers (d) S strictly contains T if S contains T but T does not contain S ; that is, if every member of T is also in S , but at least one member of S is not in T (Figure 1.3.1). (e) The complement of S , denoted by S c , is the set of elements in the universal set that are not in S . (f ) The union of S and T , denoted by S and T (Figure 1.3.1(b)). [ T , is the set of elements in at least one of S (g) The intersection of S and T , denoted by S T (Figure 1.3.1(c)). If S (Figure 1.3.1(d)). T \ D ; T , is the set of elements in both S and (the empty set), then S and T are disjoint sets \ (h) A set with only one member x0 is a singleton set, denoted by x0 . g f S T S T S T (a) S ∪ T = shaded region (b) S T S T S ∩ T = shaded region (c) S ∩ T = ∅ (d) Figure 1.3.1 Example 1.3.1 Let and x is rational ; and ˚ and x is irrational : Then S these sets are T and S S [ U , and the inclusion is strict in both cases. The unions of pairs of ˚ ˇ ˇ S; S T D U [ D S; and T U [ D S; and their intersections are S T \ D T; S U \ D U; and T U \ : D ; Also, S U D T and S T D U: Section 1.3 TheReal Line 21 Every set S contains the empty set , for to say that is not in S , which is absurd since ; is not contained in S is to say that has no members. If S is any set, some member of then ; ; ; S c .S c /c S and If S is a set of real numbers, then S R. The deﬁnitions of union and intersection have generalizations: If F is an arbitrary colS is the set of all elements that are members of at least lection of sets, then one of the sets in F , and F is the set of all elements that are members of every ˚ set in F . The union and intersection of ﬁnitely many sets S1, . . . , Sn are also written as 1 Sk. The union and intersection of an inﬁnite sequence 1 of sets g1k D 1 Sk and 1 Sk and are written as S Example 1.3.2 If F is the collection of sets ˇ ˇ 1k D n k D 1k D 1 Sk Sk =2; then ˚ =2 and S S F 2 D x 1= : Example 1.3.3 If, for each positive integer k, the set Sk is the set of real numbers 1 Sk is the set of rational that can be written as x numbers and m= k for some integer m, then 1 Sk is the set of integers. 1k D D 1k D S T Open and Closed Sets If a and b are in the extended reals and a < b, then the open interval .a; b/ is deﬁned by .a; b/ D x a < x < b : The open intervals .a; is the entire real line. 1 1 / and . ; b/ are semi-inﬁnite if a and b are ﬁnite, and . ˚ ˇ ˇ ; 1 / 1 Deﬁnition 1.3.2 If x0 is a real number and > 0, then the open interval .x0 / is an -neighborhood of x0. If a set S contains an -neighborhood of x0, then S is a neighborhood of x0, and x0 is an interior point of S (Figure 1.3.2). The set of interior points of S is the interior of S , denoted by S 0. If every point of S is an interior point (that is, S 0 S ), then S is open. A set S is closed if S c is open. ; x0 C D 22 Chapter 1 TheRealNumbers x0 − ( x0 x0 + ) S = four line segments x0 = interior point of S Figure 1.3.2 The idea of neighborhood is fundamental and occurs in many other contexts, some of which we will see later in this book. Whatever the context, the idea is the same: some deﬁnition of “closeness” is given (for example, two real numbers are “close” if their difference is “small”), and a neighborhood of a point x0 is a set that contains all points sufﬁciently close to x0. Example 1.3.4 An open interval .a; b/ is an open set, because if x0 , then min a; b x0 x0 f g .a; b/ and 2 .x0 ; x0 / C .a; b/: D / is open, and therefore The entire line R ; . 1 also open, for to deny this is to say that which is absurd because R and ; (Exercise 1.3.18). is contains a point that is not an interior point, c / is closed. Thus, are both open and closed. They are the only subsets of R with this property ; is open, R . Rc/ is closed. However, contains no points. Since . D D ; 1 ; ; ; ; A deleted neighborhood of a point x0 is a set that contains every point of some neigh- borhood of x0 except for x0 itself. For example, S D x 0 < x j x0 j < is a deleted neighborhood of x0. We also say that it is a deleted -neighborhood of x0. ˚ ˇ ˇ Theorem 1.3.3 (a) The union of open sets is open: (b) The intersection of closed sets is closed: These statements apply to arbitrary collections, ﬁnite or inﬁnite, of open and closed sets: Proof (a) Let G be a collection of open sets and S D [ G G G : 2 2 G0 for some G0 in G , and since G0 is open, it contains some If x0 S , then x0 S , this -neighborhood is in S , which is consequently a neighborhood of x0. Since G0 neighborhood of x0. Thus, S is a neighborhood of each of its points, and therefore open, by deﬁnition. 2 ˇ ˇ ˚ (b) Let F be a collection of closed sets and T F c D (Exercise 1.3.7) and, since each F c is open, T c is open, from (a). There- . Then [ fore, T is closed, by deﬁnition. 2 ˚ ˇ ˇ Section 1.3 TheReal Line 23 Example 1.3.5 If < a < b < , the set 1 1 Œa; b x a x b D is closed, since its complement is the union of the open sets . that Œa; b is a closed interval. The set 1 ˚ ˇ ˇ ; a/ and .b; /. We say 1 is a half-closed or half-open interval if Œa; b , as is a; b ˇ however, neither of these sets is open or closed. (Why not?) Semi-inﬁnite closed intervals ˇ are sets of the form ˚ Œa; / 1 D x a x and ; a . 1 D x x a ; where a is ﬁnite. They are closed sets, since their complements are the open intervals . 1 /, respectively. ; a/ and .a; 1 ˚ ˇ ˇ ˚ ˇ ˇ Example 1.3.4 shows that a set may be both open and closed, and Example 1.3.5 shows that a set may be neither. Thus, open and closed are not opposites in this context, as they are in everyday speech. Example 1.3.6 From Theorem 1.3.3 and Example 1.3.4, the union of any collection of open intervals is an open set. (In fact, it can be shown that every nonempty open subset of R is the union of open intervals.) From Theorem 1.3.3 and Example 1.3.5, the intersection of any collection of closed intervals is closed. It can be shown that the intersection of ﬁnitely many open sets is open, and that the union of ﬁnitely many closed sets is closed. However, the intersection of inﬁnitely many open sets need not be open, and the union of inﬁnitely many closed sets need not be closed (Exercises 1.3.8 and 1.3.9). Deﬁnition 1.3.4 Let S be a subset of R. Then (a) x0 is a limit point of S if every deleted neighborhood of x0 contains a point of S . (b) x0 is a boundary point of S if every neighborhood of x0 contains at least one point in S and one not in S . The set of boundary points of S is the boundary of S , denoted by @S . The closure of S , denoted by S , is S @S . S D [ S and there is a neighborhood of x0 that contains (c) x0 is an isolated point of S if x0 no ot
her point of S . 2 (d) x0 is exterior to S if x0 is in the interior of S c . The collection of such points is the exterior of S . Example 1.3.7 Let S ; . 1 D 1 [ .1; 2/ 3 g [ f . Then 24 Chapter 1 TheRealNumbers (a) The set of limit points of S is . ; 1 (b) @S ; . 1; 1; 2; 3 1 (c) 3 is the only isolated point of S . (d) The exterior of S is . and S D f 1; 1/ .2; 3/ D g 1 1 [ [ Œ1; 2. Œ1; 2 3 . g [ f [ .3; /. 1 [ Example 1.3.8 For n In 1, let 1 2n D C 1 2n ; 1 Then (a) The set of limit points of S is S (b) @S 1=n .n x x D (c) S has no isolated points. ˇ ˚ ˇ (d) The exterior of S is 0 or x D D [ f and S D 1 In: 1 n [ D . 0 g 2/ and 2n 1 C ; 2 2n Example 1.3.9 Let S be the set of rational numbers. Since every interval contains a R. rational number (Theorem 1.1.6), every real number is a limit point of S ; thus, S Since every interval also contains an irrational number (Theorem 1.1.7), every real number R. The interior and exterior of S are both empty, and is a boundary point of S ; thus @S S has no isolated points. S is neither open nor closed. D D The next theorem says that S is closed if and only if S S (Exercise 1.3.14). D Theorem 1.3.5 A set S is closed if and only if no point of S c is a limit point of S: S c. Since S c is open, there is a neighborhood Proof Suppose that S is closed and x0 of x0 that is contained in S c and therefore contains no points of S . Hence, x0 cannot be a limit point of S . For the converse, if no point of S c is a limit point of S then every point in S c must have a neighborhood contained in S c . Therefore, S c is open and S is closed. 2 Theorem 1.3.5 is usually stated as follows. Corollary 1.3.6 A set is closed if and only if it contains all its limit points: Theorem 1.3.5 and Corollary 1.3.6 are equivalent. However, we stated the theorem as we did because students sometimes incorrectly conclude from the corollary that a closed set must have limit points. The corollary does not say this. If S has no limit points, then the set of limit points is empty and therefore contained in S . Hence, a set with no limit points is closed according to the corollary, in agreement with Theorem 1.3.5. For example, any ﬁnite set is closed. More generally, S is closed if there is a ı > 0 such ı for every pair of distinct points in S . x; y j y x j f g Open Coverings Section 1.3 TheReal Line 25 A collection H of open sets is an open covering of a set S if every point in S is contained in a set H belonging to H ; that is, if S H H H . [ 2 Example 1.3.10 The sets ˚ ˇ ˇ 1; 2; : : : ; n; : : : ; g S1 S3 D D Œ0; 1; S2 1 2 1 ; : : : ; and S4 .0; 1/ D are covered by the families of open intervals H1 H2 H3 D D x n D ( n 1 C and N D positive integer), 0 < x < 1 1; 2; 2; : : : ; ) H4 .0 respectively. Theorem 1.3.7 (Heine–Borel Theorem) If H is an open covering of a closed H consisting of ﬁnitely and bounded subset S of the real line; then S has an open covering many open sets belonging to H : Proof Since S is bounded, it has an inﬁmum ˛ and a supremum ˇ, and, since S is closed, ˛ and ˇ belong to S (Exercise 1.3.17). Deﬁne e St S \ D Œ˛; t for t ˛; and let F t ˛ t D ˇ and ﬁnitely many sets from H cover St : Since Sˇ the completeness of the reals. D ˇ ˇ ˚ S , the theorem will be proved if we can show that ˇ F . To do this, we use 2 2 Since ˛ S , S˛ is the singleton set H because H covers S ; therefore, ˛ it has a supremum . First, we wish to show that it sufﬁces to rule out the possibility that < ˇ. We consider two cases. , which is contained in some open set H˛ from ˛ f F . Since F is nonempty and bounded above by ˇ, ˇ by deﬁnition of F , ˇ. Since D 2 g 26 Chapter 1 TheRealNumbers CASE 1. Suppose that < ˇ and of S (Theorem 1.3.5). Consequently, there is an > 0 such that 62 S . Then, since S is closed, is not a limit point Œ ; S \ C ; D ; so S D from H , while S S C C . However, the deﬁnition of implies that S does not. This is a contradiction. has a ﬁnite subcovering Now we know that CASE 2. Suppose that < ˇ and 2 contains and, along with , an interval Œ a ﬁnite covering f H1; : : : ; Hn; H g H1; : : : ; Hn . This contradicts the deﬁnition of . S . Then there is an open set H in H that for some positive . Since S ; has of sets from H , it follows that S has the ﬁnite covering C C g f contains ˇ and along with ˇ, an interval of the form Œˇ Since Sˇ ﬁnite collection is covered by a ﬁnite collection of sets H1; : : : ; Hk; Hˇ . Since Sˇ D ˇ, which is in S . Therefore, there is an open set Hˇ in H that , for some positive . , Sˇ is covered by the ; ˇ C H1; : : : ; Hk g f D Henceforth, we will say that a closed and bounded set is compact. The Heine–Borel theorem says that any open covering of a compact set S contains a ﬁnite collection that also covers S . This theorem and its converse (Exercise 1.3.21) show that we could just as well deﬁne a set S of reals to be compact if it has the Heine–Borel property; that is, if every open covering of S contains a ﬁnite subcovering. The same is true of Rn, which we study in Section 5.1. This deﬁnition generalizes to more abstract spaces (called topological spaces) for which the concept of boundedness need not be deﬁned. g S , we are ﬁnished. f 2N 1/ cover S1. Example 1.3.11 Since S1 in Example 1.3.10 is compact, the Heine–Borel theorem implies that S1 can be covered by a ﬁnite number of intervals from H1. This is easily veriﬁed, since, for example, the 2N intervals from H1 centered at the points xk k The Heine–Borel theorem does not apply to the other sets in Example 1.3.10 since they are not compact: S2 is unbounded and S3 and S4 are not closed, since they do not contain all their limit points (Corollary 1.3.6). The conclusion of the Heine–Borel theorem does not hold for these sets and the open coverings that we have given for them. Each point in S2 is contained in exactly one set from H2, so removing even one of these sets leaves a point of S2 uncovered. If H 3 is any ﬁnite collection of sets from H3, then k=2N .0 D e 1 n 62 [ H H H 3 2 for n sufﬁciently large. Any ﬁnite collection interval .0; max/, where ˚ ˇ .0; 1/; : : : ; .0; n/ ˇ g f e from H4 covers only the max max f D 1; : : : ; n < 1: g The Bolzano–Weierstrass Theorem As an application of the Heine–Borel theorem, we prove the following theorem of Bolzano and Weierstrass. Theorem 1.3.8 (Bolzano–Weierstrass Theorem) Every bounded inﬁnite set of real numbers has at least one limit point: Section 1.3 TheReal Line 27 Proof We will show that a bounded nonempty set without a limit point can contain only a ﬁnite number of points. If S has no limit points, then S is closed (Theorem 1.3.5) and every point x of S has an open neighborhood Nx that contains no point of S other than x. The collection H Nx x S 2 D is an open covering for S . Since S is also bounded, Theorem 1.3.7 implies that S can be ˇ ˇ covered by a ﬁnite collection of sets from H , say Nx1, . . . , Nxn. Since these sets contain only x1, . . . , xn from S , it follows that S x1; : : : ; xn ˚ . D f g 1.3 Exercises 1. Find S (a) S \ T , .S \ .0; 1/, T (c) S 2. Let Sk D D ; . 1 .1 D D /, T 1 1= k; 2 C D ; T /, S [ (b) S T , .S x D T /c , and S c T c. [ x2 > 4 , T x2 < 9 [ x D / .1; ˇ ˚ ˇ 1 (d. Find 1= k, k 1/, T D 1 (a) Sk (b) 1 Sk (c) S c k (d) 1 S c k 1 [k D 1 \k D 1 \k D 1 1 [k D 3. 4. Prove: If A and B are sets and there is a set X such that A B. A X, then A X B X [ D B [ X and \ D \ D Find the largest such that S contains an -neighborhood of x0. (a) x0 (b) x0 1 D 3 4 , S 5, S D 1 2 ; 1 1, S D 2 2 ; 3 .0; 2/ (d) x0 D 5. Describe the following sets as open, closed, or neither, and ﬁnd S 0, .S c /0, and D D D (c) x0 .S 0/c . (a) S (b) S 1; 2/ . Œ D [ (d) S 2 [ D T c and .S T /c 6. ˚ 7. Let F be a collection of sets and deﬁne (c) S D Prove that .S D D 3; \ [ Œ3; / 1 Œ7; 8 S c ; 1/ . 1 x x D T /c ˇ [ ˇ / 1 .2; [ integer and U F F F : 2 D [ Prove that (a) I c 2 2 (a) Show that the intersection of ﬁnitely many open sets is open. and (b. ˇ ˇ 2 F ˇ ˇ 28 Chapter 1 TheRealNumbers (b) Give an example showing that the intersection of inﬁnitely many open sets may fail to be open. 9. (a) Show that the union of ﬁnitely many closed sets is closed. (b) Give an example showing that the union of inﬁnitely many closed sets may fail to be closed. 10. Prove: D n i 1 Ui . V , then V is a neighborhood of x0. (a) If U is a neighborhood of x0 and U (b) If U1, . . . , Un are neighborhoods of x0, so is Find the set of limit points of S , @S , S , the set of isolated points of S , and the exterior of S . (a) S (b) S (c) S (d) S Prove: A limit point of a set S is either an interior point or a boundary point of S . Prove: An isolated point of S is a boundary point of S c . ; . [ 1 all integers g .n; n n 1/ C 1=n integer 1; 2; 3; : : : D D / 1 .2; 3/ g [ [ f .7; T 2/ D ˇ ˇ ˇ ˇ 4 ˚ Prove: (a) A boundary point of a set S is either a limit point or an isolated point of S . (b) A set S is closed if and only if S D Prove or disprove: A set has no limit points if and only if each of its points is isolated. (a) Prove: If S is bounded above and ˇ (b) State the analogous result for a set bounded below. sup S , then ˇ @S . S . D 2 11. 12. 13. 14. 15. 16. Prove: If S is closed and bounded, then inf S and sup S are both in S . If a nonempty subset S of R is both open and closed, then S 17. 18. 19. Let S be an arbitrary set. Prove: (a) @S is closed. (b) S 0 is open. (c) The exterior R. D of S is open. (d) The limit points of S form a closed set. (e) S S . D 20. Give counterexamples to the following false statements. (a) The isolated points of a set form a closed set. (b) Every open set contains at least two points. (c) If S1 and S2 are arbitrary sets, then @.S1 (d) If S1 and S2 are arbitrary sets, then @.S1 (e) The supremum of a bounded nonempty set is the greatest of its limit points. (f ) If S is any set, then @.@S / (g) If S is any set, then @S (h) If S1 and S2 are arbitrary sets, then .S1 @S2. @S2. @S1 @S1 S2/ S2/ D @S . D D S2/0 [ \ [ \ @ Section 1.3 TheReal Line 29 21. Let S be a nonempty subset of
R such that if H is any open covering of S , then S H comprised of ﬁnitely many open sets from H . Show that has an open covering S is compact. e 22. A set S is. in a set T if S T S . (a) Prove: If S and T are sets of real numbers and S T , then S is dense in T if and only if every neighborhood of each point in T contains a point from S . (b) State how (a) shows that the deﬁnition given here is consistent with the re- stricted deﬁnition of a dense subset of the reals given in Section 1.1. 23. Prove: (a) .S1 Prove: 24. S2/b) S 0 1 [ S 0 2 .S1 [ S2/0 (a) @.S1 S2/ @S1 @S2 [ (b) @.S1 @S1 @S2 [ S2/ \ @S c D (d) @S (c) @S [ @S (e) @.S T / @S @T [ CHAPTER 2 Diﬀerential Calculus of Functions of One Variable , and monotonic functions. IN THIS CHAPTER we study the differential calculus of functions of one variable. SECTION 2.1 introduces the concept of function and discusses arithmetic operations on functions, limits, one-sided limits, limits at ˙1 SECTION 2.2 deﬁnes continuity and discusses removable discontinuities, composite functions, bounded functions, the intermediate value theorem, uniform continuity, and additional properties of monotonic functions. SECTION 2.3 introduces the derivative and its geometric interpretation. Topics covered include the interchange of differentiation and arithmetic operations, the chain rule, one-sided derivatives, extreme values of a differentiable function, Rolle’s theorem, the intermediate value theorem for derivatives, and the mean value theorem and its consequences. SECTION 2.4 presents a comprehensive discussion of L’Hospital’s rule. SECTION 2.5 discusses the approximation of a function f by the Taylor polynomials of f and applies this result to locating local extrema of f . The section concludes with the extended mean value theorem, which implies Taylor’s theorem. 2.1 FUNCTIONS AND LIMITS In this section we study limits of real-valued functions of a real variable. You studied limits in calculus. However, we will look more carefully at the deﬁnition of limit and prove theorems usually not proved in calculus. A rule f that assigns to each member of a nonempty set D a unique member of a set Y is a function from D to Y . We write the relationship between a member x of D and the member y of Y that f assigns to x as f .x/: y D The set D is the domain of f , denoted by Df . The members of Y are the possible values of f . If y0 Y and there is an x0 in D such that f .x0/ y0 then we say that f attains 2 30 D Section 2.1 FunctionsandLimits 31 or assumes the value y0. The set of values attained by f is the range of f . A real-valued function of a real variable is a function whose domain and range are both subsets of the reals. Although we are concerned only with real-valued functions of a real variable in this section, our deﬁnitions are not restricted to this situation. In later sections we will consider situations where the range or domain, or both, are subsets of vector spaces. Example 2.1.1 The functions f , g, and h deﬁned on . f .x/ D x2; g.x/ D sin x; and have ranges Œ0; /, Œ 1 1; 1, and .0; /, respectively. 1 Example 2.1.2 The equation ; 1 h.x/ / by 1 ex D does not deﬁne a function except on the singleton set (2.1.1), while if x > 0, two real numbers satisfy (2.1.1). However, the conditions . If x < 0, no real number satisﬁes 0 f g Œf .x/2 x D (2.1.1) deﬁne a function f on Df Œ0; / with values f .x/ Œf .x/2 D x and f .x/ 0 D 1 Œg.x/2 x and g.x/ 0 D px. Similarly, the conditions D deﬁne a function g on Dg Œ0; Œ0; ; 0, respectively. / and . D 1 1 1 / with values g.x/ px. The ranges of f and g are D D ; 1 / and values f .x/ It is important to understand that the deﬁnition of a function includes the speciﬁcation of its domain and that there is a difference between f , the name of the function, and f .x/, the value of f at x. However, strict observance of these points leads to annoying verbosity, such as “the function f with domain . x.” We will avoid this 1 in two ways: (1) by agreeing that if a function f is introduced without explicitly deﬁning Df , then Df will be understood to consist of all points x for which the rule deﬁning f .x/ makes sense, and (2) by bearing in mind the distinction between f and f .x/, but not emphasizing it when it would be a nuisance to do so. For example, we will write “consider 1; 1 the function f .x/ p1 1= sin x,” rather than “consider by f .x/ 1= sin x.” We will also write the function g deﬁned for x c for all x. f x2,” rather than “consider the function f deﬁned on Œ c (constant) to denote the function f deﬁned by f .x/ D x2,” or “consider the function g.x/ D integer) by g.x/ D Our deﬁnition of function is somewhat intuitive, but adequate for our purposes. Moreover, it is the working form of the deﬁnition, even if the idea is introduced more rigorously to begin with. For a more precise deﬁnition, we ﬁrst deﬁne the Cartesian product X Y of two nonempty sets X and Y to be the set of all ordered pairs .x; y/ such that x X and y Y ; thus, k (k p1 x; y/ x X; y Y : 2 2 ˚ ˇ ˇ 32 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Y is a function if no x in X occurs more than once as a ﬁrst A nonempty subset f of X member among the elements of f . Put another way, if .x; y/ and .x; y1/ are in f , then y1. The set of x’s that occur as ﬁrst members of f is the of f . If x is in the domain y f is the value of f at x, and we write of f , then the unique y in Y such that .x; y/ y f .x/. The set of all such values, a subset of Y , is the range of f . D 2 D Arithmetic Operations on Functions Deﬁnition 2.1.1 If Df by Dg \ ; then f g; f C ¤ ; g; and fg are deﬁned on Df Dg \ and .f .f C g/.x/ g/.x/ .fg/.x/ D D D f .x/ f .x/ g.x/; g.x/; C f .x/g.x/: The quotient f =g is deﬁned by f g .x/ D f .x/ g.x/ for x in Df \ Dg such that g.x/ 0: ¤ Example 2.1.3 If f .x/ Dg /; so f g; f Œ1; p4 px x2 and g.x/ D g; and fg are deﬁned on Df D D 1 C 1; then Df Dg D Œ1; 2 by D \ Œ 2; 2 and .f .f C g/.x/ g/.x/ p4 p4 x2 x2 px px 1; 1; C D D and .fg/.x/ .p4 D x2/.px 1/ D .4 x2/.x 1/: (2.1.2) The quotient f =g is deﬁned on .1; 2 by p f g .x/ x2 1 4 x : D r Although the last expression in (2.1.2) is also deﬁned for represent fg for such x; since f and g are not deﬁned on . 1 ; 1 < x < 2. 2; it does not Example 2.1.4 If c is a real number, the function cf deﬁned by .cf /.x/ cf .x/ can be regarded as the product of f and a constant function. Its domain is Df . The sum and product of n . 2/ functions f1, . . . , fn are deﬁned by D .f1 f2 C C C fn/.x/ f1.x/ f2.x/ C D C C fn.x/ Section 2.1 FunctionsandLimits 33 .f1f2 fn/.x/ D f1.x/f2.x/ fn.x/ (2.1.3) 1 Dfi , provided that D is nonempty. If f1 f2 D D D fn, then (2.1.3) and on D D deﬁnes the nth power of f : D n i T .f n/.x/ D .f .x//n: From these deﬁnitions, we can build the set of all polynomials p.x/ a0 C D a1x C C anxn; starting from the constant functions and f .x/ rational function D C C The domain of r is the set of points where the denominator is nonzero. C C C C D ¤ r .x/ .bm 0/: a0 b0 a1x b1x anxn bmxm x. The quotient of two polynomials is a Limits The essence of the concept of limit for real-valued functions of a real variable is this: If L L means that the value f .x/ can be made as close is a real number, then limx to L as we wish by taking x sufﬁciently close to x0. This is made precise in the following deﬁnition. x0 f .xx) x x0 − δ x0 x0 + δ Figure 2.1.1 34 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Deﬁnition 2.1.2 We say that f .x/ approaches the limit L as x approaches x0, and write if f is deﬁned on some deleted neighborhood of x0 and, for every > 0, there is a ı > 0 such that lim x0 x ! f .x/ L; D if f .x/ j < L j Figure 2.1.1 depicts the graph of a function for which limx 0 < x x0 < ı: j j (2.1.4) (2.1.5) x0 f .x/ exists. ! Example 2.1.5 If c and x are arbitrary real numbers and f .x/ cx, then D To prove this, we write If c ¤ 0, this yields if lim x0 x ! f .x/ cx0: D f .x/ j cx0 cx cx0 c x jj x0 : j j D j j D j f .x/ j cx0 j < (2.1.6) j where ı is any number such that 0 < ı so (2.1.6) holds for all x. x x0 j c = j < ı; . If c j D 0, then f .x/ cx0 D 0 for all x, We emphasize that Deﬁnition 2.1.2 does not involve f .x0/, or even require that it be deﬁned, since (2.1.5) excludes the case where x x0. D Example 2.1.6 If then f .x/ x sin D 1 x ; 0; x ¤ even though f is not deﬁned at x0 0 f .x/ lim D x 0 ! 0, because if D 0 < < ı x j j D ; then On the other hand, the function f .x/ j 0 j D x sin ˇ ˇ ˇ ˇ sin ; ; ¤ g.x/ D has no limit as x approaches 0, since it assumes all values between neighborhood of the origin (Exercise 2.1.26). 1 and 1 in every The next theorem says that a function cannot have more than one limit at a point. Section 2.1 FunctionsandLimits 35 Theorem 2.1.3 If limx x0 f .x/ exists; then it is unique that is; if I ! f .x/ lim x0 x ! L1 and D lim x0 x ! f .x/ L2; D (2.1.7) then L1 L2: D Proof Suppose that (2.1.7) holds and let > 0. From Deﬁnition 2.1.2, there are positive numbers ı1 and ı2 such that f .x/ j Li j < if 0 < x j x0 j < ıi ; i 1; 2: D If ı D min.ı1; ı2/, then L1 L2 f .x/ L2 j C j C j We have now established an inequality that does not depend on x; that is, j D j j f .x/ j L2 < 2 0 < if x j j x0 < ı: j L1 L1 f .x/ f .x/ j Since this holds for any positive , L1 < 2: L2 L2. j L1 D Deﬁnition 2.1.2 is not changed by replacing (2.1.4) with f .x/ j L j < K; (2.1.8) where K is a positive constant, because if either of (2.1.4) or (2.1.8) can be made to hold for any > 0 by making sufﬁciently small and positive, then so can the other (Exercise 2.1.5). This may seem to be a minor point, but it is often convenient to work with (2.1.8) rather than (2.1.4), as we will see in the proof of the following theorem. x0 x j j A Useful Theorem about Limits Theorem 2.1.4 If lim x0 x ! f .x/ D L1 and lim x0 x ! g.x/ L2; D then and, if L2 0, ¤ .f .f lim x0 x ! lim x0 x ! lim x0 x ! g/.x/ C g/.x/ .fg/.x/ D D D L1 C L2; L2; L1 L1L2; lim x0 x ! f g .x/ D L1 L2 : (2.1.9) (2.1.10) (2.1.11) (2.1.12) (2.1.13) (2.1.14) 36 Chap
ter 2 DifferentialCalculusofFunctionsofOneVariable Proof From (2.1.9) and Deﬁnition 2.1.2, if > 0, there is a ı1 > 0 such that f .x/ < ı1, and a ı2 > 0 such that j < L1 j if 0 < x j x0 j g.x/ j L2 j < if 0 < x j x0 j < ı2. Suppose that so that (2.1.15) and (2.1.16) both hold. Then 0 < x j x0 j < ı D min.ı1; ı2/; (2.1.15) (2.1.16) (2.1.17) .f j ˙ g/.x/ .L1 ˙ L2/ j D j j .f .x/ f .x/ L1/ L1 ˙ j C j which proves (2.1.10) and (2.1.11). To prove (2.1.12), we assume (2.1.17) and write .g.x/ g.x/ L2/ j < 2; L2 j .fg/.x/ j L1L2 j D j D j f .x/g.x/ L1L2 j C L2 L2/ L2 / j L2.f .x/ f .x/ L2 L1/ j L1 j C j jj (from (2.1.15) and (2.1.16)) j L1 / L2 / j C j j from (2.1.15) j C j L2 f .x/.g.x/ f .x/ g.x/ j C j L1 jj f .x/ f .x/ j . j . j . .1 C j C j L1 L1 j C j j C j L2 / j j if < 1 and x satisﬁes (2.1.17). This proves (2.1.12). To prove (2.1.14), we ﬁrst observe that if L2 0, there is a ı3 > 0 such that ¤ so if g.x/ j L2 j < j L2 2 j ; g.x/ j j > j L2 2 j (2.1.18) To see this, let L L2 and D 0 < x j x0 j < ı3: =2 in (2.1.4). Now suppose that L2 j x j x0 j < min.ı1; ı2; ı3/; D j 0 < so that (2.1.15), (2.1.16), and (2.1.18) all hold. Then Section 2.1 FunctionsandLimits 37 f g .x/ ˇ ˇ ˇ ˇ L1 L2 x/ g.x/ L2f .x/ L1 L2 ˇ ˇ L1g.x/ ˇ j ˇ g.x/L2 j j L2f .x/ 2 j L1g.x/ j 2 L2 2 L2 2 L2 2 L2 j j j j j j j j D This proves (2.1.14). L2Œf .x/ L1 C L1ŒL2 g.x/ j 2 j (from (2.1.18)) L2 Œ j jj 2 f .x/ L1 L1 L2 jj g.x/ j j C j L2 . j 2 L1 / j j C j (from (2.1.15) and (2.1.16)): Successive applications of the various parts of Theorem 2.1.4 permit us to ﬁnd limits without the –ı arguments required by Deﬁnition 2.1.2. Example 2.1.7 Use Theorem 2.1.4 to ﬁnd lim 2 x ! 9 x x2 1 C and .9 lim 2 x ! x2/.x 1/: C Solution If c is a constant, then limx x0. Therefore, from Theorem 2.1.4, x0 c ! D c, and, from Example 2.1.5, limx x0 x ! D x2/ .9 lim 2 x ! D D D 9 lim x 2 ! lim 2 x ! 22 9 9 D lim x 2 ! . lim 2 x ! 5; x2 x/2 and Therefore, and .x lim 2 x ! C 1/ D x lim 2 x ! C 1 lim 2 x ! 1 2 C D D 3: lim 2 x ! 9 x x2 1 D C .9 lim x 2 ! lim 2 x ! .x x2/ 1/ D C 5 3 .9 lim 2 x ! x2/.x 1/ C D .9 lim 2 x ! x2/ lim 2 x ! .x 1/ 3 5 D D C 15: One-Sided Limits The function f .x/ D 2x sin px 38 Chapter 2 DifferentialCalculusofFunctionsofOneVariable satisﬁes the inequality f .x/ j j < if 0 < x < ı 0, since f is not deﬁned for negative x, as it must be to satisfy the conditions of Deﬁnition 2.1.2 with x0 =2. However, this does not mean that limx 0. The function 0 and L 0 f .x/ D D ! D D g.x; x ¤ can be rewritten as g.x/ hence, every open interval containing x0 g.x2/ g.x1/ j j (Exercise 2.1.26). x x 1; x > 0; x < 0 1; D I C D is as close to 2 as we please. Therefore, limx 0 also contains points x1 and x2 such that x0 g.x/ does not exist ! Although f .x/ and g.x/ do not approach limits as x approaches zero, they each exhibit a deﬁnite sort of limiting behavior for small positive values of x, as does g.x/ for small negative values of x. The kind of behavior we have in mind is deﬁned precisely as follows. y µ λ y = f (x) x0 x f (x) = λ lim x x0 − f (x) = µ lim x x0 + Figure 2.1.2 Deﬁnition 2.1.5 (a) We say that f .x/ approaches the left-hand limit L as x approaches x0 from the left, and write lim x0 ! x f .x/ L; D if f is deﬁned on some open interval .a; x0/ and, for each > 0, there is a ı > 0 such that f .x/ j L j < if x0 ı < x < x0: Section 2.1 FunctionsandLimits 39 (b) We say that f .x/ approaches the right-hand limit L as x approaches x0 from the right, and write lim x0 ! x C f .x/ L; D if f is deﬁned on some open interval .x0; b/ and, for each > 0, there is a ı > 0 such that f .x/ j L j < if x0 < x < x0 ı: C Figure 2.1.2 shows the graph of a function that has distinct left- and right-hand limits at a point x0. Example 2.1.8 Let f .x/ D If x < 0, then f .x/ x=x D 1, so D ; x x j j 0: x ¤ If x > 0, then f .x/ x=x D D 1, so lim 0 ! x f .x/ 1: D lim 0 C ! x f .x/ 1: D Example 2.1.9 Let g.x/ x C j x .1 j x C x/ sin 1 x ; D 0: x ¤ If x < 0, then so since g.x/ x sin D 1 x ; lim 0 ! x g.x/ 0; D g.x sin if < x < 0; that is, Deﬁnition 2.1.5(a) is satisﬁed with ı . If x > 0, then D g.x/ .2 C D x/ sin 1 x ; 2 and 2 in every interval .0; ı/. Hence, g.x/ does not which takes on every value between approach a right-hand limit at x approaches 0 from the right. This shows that a function may have a limit from one side at a point but fail to have a limit from the other side. 40 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Example 2.1.10 We leave it to you to verify that sin px x x lim 0 C ! x lim 0 ! lim 0 C ! x 1; D 1; D 0; D and limx 0 ! sin px does not exist. Left- and right-hand limits are also called one-sided limits. We will often simplify the notation by writing lim x0 ! x f .x/ f .x0 / D and lim x0 ! x C f .x/ f .x0 /: C D The following theorem states the connection between limits and one-sided limits. We leave the proof to you (Exercise 2.1.12). Theorem 2.1.6 A function f has a limit at x0 if and only if it has left- and right-hand limits at x0; and they are equal. More speciﬁcally; if and only if lim x0 x ! f .x/ L D f .x0 / C D f .x0 / D L: With only minor modiﬁcations of their proofs (replacing the inequality 0 < ı < x < x0 or x0 < x < x0 by x0 C rems 2.1.3 and 2.1.4 remain valid if “limx throughout (Exercise 2.1.13). ! < ı ı), it can be shown that the assertions of Theo” x0 ” is replaced by “limx ” or “limx x0 x0 x0 x j j ! ! C Limits at ˙1 Limits and one-sided limits have to do with the behavior of a function f near a limit point of Df . It is equally reasonable to study f for large positive values of x if Df is unbounded above or for large negative values of x if Df is unbounded below. Deﬁnition 2.1.7 We say that f .x/ approaches the limit L as x approaches write , and 1 if f is deﬁned on an interval .a; / and, for each > 0, there is a number ˇ such that lim x !1 f .x/ L; D 1 f .x/ j L j < if x > ˇ: Figure 2.1.3 provides an illustration of the situation described in Deﬁnition 2.1.7. Section 2.1 FunctionsandLimits 41 y L + L L − f (x) = L lim x ∞ x β Figure 2.1.3 We leave it to you to deﬁne the statement “limx f .x/ to show that Theorems 2.1.3 and 2.1.4 remain valid if x0 is replaced throughout by !1 D L” (Exercise 2.1.14) and or 1 (Exercise 2.1.16). 1 Example 2.1.11 Let f .x/ 1 D 1 x2 ; g.x and h.x/ sin x: D Then since and since j lim x !1 f .x/ 1; D f .x/ j 1 j D 1 x2 < if x > 1 p ; lim x !1 g.x/ 2; D 2x 2 g.x.x/ does not exist, since h assumes all values between C < < if and 1 in any However, limx semi-inﬁnite interval .; !1 ˇ ˇ ˇ ˇ /. 1 We leave it to you to show that limx !1 h.x/ does not exist (Exercise 2.1.17). limx !1 f .x/ D 1, limx !1 g.x/ 2, and D 42 Chapter 2 DifferentialCalculusofFunctionsofOneVariable We will sometimes denote limx respectively. f .x/ and limx !1 !1 f .x/ by f . / and f . 1 /, 1 Inﬁnite Limits The functions and f .x/ 1 x ; D g.x/ 1 x2 D ; p.x/ sin D 1 x ; do not have limits, or even one-sided limits, at x0 ways: D q.x/ 1 x2 sin D 1 x 0. They fail to have limits in different f .x/ increases beyond bound as x approaches 0 from the right and decreases beyond bound as x approaches 0 from the left; g.x/ increases beyond bound as x approaches zero; p.x/ oscillates with ever-increasing frequency as x approaches zero; q.x/ oscillates with ever-increasing amplitude and frequency as x approaches 0. The kind of behavior exhibited by f and g near x0 simple to lead us to deﬁne inﬁnite limits. 0 is sufﬁciently common and D Deﬁnition 2.1.8 We say that f .x/ approaches and write 1 as x approaches x0 from the left, f .x/ lim x0 ! x D 1 or f .x0 / ; D 1 if f is deﬁned on an interval .a; x0/ and, for each real number M , there is a ı > 0 such that f .x/ > M if x0 ı < x < x0: Example 2.1.12 We leave it to you to deﬁne the other kinds of inﬁnite limits (Exercises 2.1.19 and 2.1.21) and show that 1 ; x D 1 1 x2 D x2 x lim 0 ! C lim !1 D x lim 0 x ! lim 0 ! lim x !1 x lim 0 C ! x 1 x2 D x2 D 1I 1 x D 1I 1 x2 D 1I lim 0 x ! and lim x !1 x3 ; D 1 lim !1 x x3 : D 1 Section 2.1 FunctionsandLimits 43 Throughout this book, “limx x0 f .x/ exists” will mean that ! lim x0 x ! f .x/ D L; where L is ﬁnite. To leave open the possibility that L , we will say that D ˙1 f .x/ exists in the extended reals. lim x0 x ! This convention also applies to one-sided limits and limits as x approaches We mentioned earlier that Theorems 2.1.3 and 2.1.4 remain valid if “limx . ˙1 x0 ” is replaced by “limx . ˙1 Moreover, the counterparts of (2.1.10), (2.1.11), and (2.1.12) in all these versions of Theorem 2.1.4 remain valid if either or both of L1 and L2 are inﬁnite, provided that their right sides are not indeterminate (Exercises 2.1.28 and 2.1.29). Equation (2.1.14) and its counterparts remain valid if L1=L2 is not indeterminate and L2 .” They are also valid with x0 replaced by 0 (Exercise 2.1.30). ” or “limx x0 x0 ! ! ! C ¤ Example 2.1.13 Results like Theorem 2.1.4 yield sinh x lim x !1 sinh x lim ! lim x !1 and ex x e 2 1 2 D 0/ ex ; D 1 x e 2 1 2 D lim x !1 1 . 2 1 lim x !1 1 .0 2 / 1 ; D 1 x e lim x !1 lim x !1 x D 0 1 0: D lim x !1 ex lim x !1 x e ex lim !1 x lim !1 x e x Example 2.1.14 If we cannot obtain limx !1 f .x/ D f .x/ by writing e2x ex; because this produces the indeterminate form f .x/ lim x !1 lim x !1 D e2x ex; lim x !1 . However, by writing we ﬁnd that f .x/ D 1 1 e e2x.1 x/; f .x/ lim x !1 D lim x !1 e2x 1 lim x !1 lim x !1 x e .1 D 1 0/ : D 1 44 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Example 2.1.15 Let g.x/ D 2x2 3x2 1 1 : x 2x C C Trying to ﬁnd limx written leads to an indeterminate form (try it!). However, by rewriting it as g.x/ by applying a version of Theorem 2.1.4 to this fraction as it is !1 g.x/ D 1=x 2=x 2 3 C C 1=x2 1=x2 ; 0; x ¤ we ﬁnd that g.x/ lim x !1 D 2 3 lim x !1 lim x !1 lim x !1 lim x !1 C 1=x 2=x lim x !1 lim x !1 C 1=x2 1=x2 : Monotonic Function A function f is nondecreasing on an interval I if f .x1/ f .x2/ whenever x1 and x2 are in I and x1 < x2; (2.1.19) or nonincreasing on I if f .x1/ f .x2/ whenever x1 and x2 are in I
and x1 < x2: (2.1.20) In either case, f is on I . If can be replaced by > in (2.1.20), f is decreasing on I . In either of these two cases, f is strictly monotonic on I . can be replaced by < in (2.1.19), f is increasing on I . If Example 2.1.16 The function f .x/ x; 0 x < 1; 1 x 2; D ( 2; is nondecreasing on I D Œ0; 2 (Figure 2.1.4), and f is nonincreasing on I Œ0; 2. D y 2 1 1 2 x Section 2.1 FunctionsandLimits 45 The function g.x/ D x2 is increasing on Œ0; / (Figure 2.1.5), 1 Figure 2.1.4 y y = x 2 x Figure 2.1.5 and h.x/ x3 is decreasing on . D ; 1 / (Figure 2.1.6). 1 y x y = − x 3 Figure 2.1.6 46 Chapter 2 DifferentialCalculusofFunctionsofOneVariable In the proof of the following theorem, we assume that you have formulated the deﬁnitions called for in Exercise 2.1.19. Theorem 2.1.9 Suppose that f is monotonic on .a; b/ and deﬁne ˛ inf a<x<b D f .x/ and ˇ sup a<x<b D f .x/: (a) If f is nondecreasing; then f .a (b) If f is nonincreasing; then f .a .Here a if a C D 1 D 1 (c) If a < x0 < b, then f .x0 / and f .x0 / C / C and b D ˇ and f .b ˛ and f .b D ˇ: / / D :/ D 1 / exist and are ﬁnite if b ˛: D D 1 if f is nondecreasing; and f .x0 / C f .x0/ f .x0 / C if f is nonincreasing: f .x0 / f .x0/ f .x0 / C moreover; I Proof (a) We ﬁrst show that f .a ˛. If M > ˛, there is an x0 in .a; b/ such that f .x0/ < M . Since f is nondecreasing, , let . If ˛ > C D / / f .x/ < M if a < x < x0. Therefore, if ˛ , where > 0. Then ˛ M ˛ f .x/ < ˛ , then f .a , so C D 1 1 D 1 C D C < if a < x < x0: (2.1.21) If a D 1 equivalent to , this implies that f . ˛. If a > , let ı 1 D f .x/ < if a < x < a ˛ j C a. Then (2.1.21) is x0 D ı; f .x/ j ˛ j / 1 which implies that f .a C We now show that f .b j / ˛. D / D Since f is nondecreasing, f .x/ > M if x0 < x < b. Therefore, if ˇ f .b , where > 0. Then ˇ < f .x/ . If ˇ < , let M ˇ ˇ. If M < ˇ, there is an x0 in .a; b/ such that f .x0/ > M . , then D 1 ˇ, so / D 1 1 < if x0 < x < b: (2.1.22) If b equivalent to D 1 , this implies that f . ˇ. If b < , let ı 1 b D D x0. Then (2.1.22) is D f .x/ j ˇ j / 1 f .x/ j < if b ˇ j ı < x < b; which implies that f .b / D ˇ. (b) The proof is similar to the proof of (a) (Exercise 2.1.34). (c) Suppose that f is nondecreasing. Applying (a) to f on .a; x0/ and .x0; b/ sepa- rately shows that f .x0 / D sup a<x<x0 f .x/ and f .x0 / C D inf x0<x<b f .x/: Section 2.1 FunctionsandLimits 47 However, if x1 < x0 < x2, then hence, f .x1/ f .x0/ f .x2/ I / We leave the case where f is nonincreasing to you (Exercise 2.1.34). f .x0/ f .x0 f .x0 C /: Limits Inferior and Superior We now introduce some concepts related to limits. We leave the study of these concepts mainly to the exercises. We say that f is bounded on a set S if there is a constant M < for all x in S . such that f .x/ j j M 1 Deﬁnition 2.1.10 Suppose that f is bounded on Œa; x0/, where x0 may be ﬁnite or For a x < x0, deﬁne . 1 and Sf .x If .x I I x0/ D x sup t <x0 f .t/ x0/ D x inf t <x0 f .t/: Then the left limit superior of f at x0 is deﬁned to be and the left limit inferior of f at x0 is deﬁned to be lim x0 ! x f .x/ D x lim x0 ! Sf .x x0/; I f .x/ D x lim x0 ! If .x x0/: I lim x0 ! x .) (If x0 , we deﬁne x0 D 1 D 1 Theorem 2.1.11 If f is bounded on Œa; x0/; then ˇ the unique real number with the following properties (a) If > 0, there is an a1 in Œa; x0/ such that W limx x0 ! D f .x/ exists and is f .x/ < ˇ if a1 x < x0: C (2.1.23) (b) If > 0 and a1 is in Œa; x0/; then f .x/ > ˇ for some x Œa1; x0/: 2 Proof Since f is bounded on Œa; x0/, Sf .x I Œa; x0/. By applying Theorem 2.1.9(b) to Sf .x I Therefore, if > 0, there is an a in Œa; x0/ such that x0/ is nonincreasing and bounded on x0/, we conclude that ˇ exists (ﬁnite). ˇ =2 < Sf .x x0/ < ˇ I C =2 if a x < x0: (2.1.24) 48 Chapter 2 DifferentialCalculusofFunctionsofOneVariable I x0/ is an upper bound of t < x0 Since Sf .x the second inequality in (2.1.24) implies (2.1.23) with a1 (b), let a1 be given and deﬁne x1 implies that x0/. Therefore, a. This proves (a). To prove max.a1; a/. Then the ﬁrst inequality in (2.1.24) , f .x/ Sf .x ˚ D f .t/ D x ˇ ˇ I x0/ is the supremum of Sf .x1 x0/ > ˇ =2: I f .t/ x1 < t < x0 (2.1.25) , there is an x in Œx1; x0/ such Since Sf .x1 that I ˇ ˚ f .x/ > Sf .x1 ˇ =2: x0/ I This and (2.1.25) imply that f .x/ > ˇ . Since x is in Œa1; x0/, this proves (b). Now we show that there cannot be more than one real number with properties (a) and (b). Suppose that ˇ1 < ˇ2 and ˇ2 has property (b); thus, if > 0 and a1 is in Œa; x0/, ˇ1, we see that there there is an x in Œa1; x0/ such that f .x/ > ˇ2 is an x in Œa1; b/ such that . Letting ˇ2 D f .x/ > ˇ2 .ˇ2 ˇ1/ D ˇ1; so ˇ1 cannot have property (a). Therefore, there cannot be more than one real number that satisﬁes both (a) and (b). The proof of the following theorem is similar to this (Exercise 2.1.35). Theorem 2.1.12 If f is bounded on Œa; x0/; then ˛ the unique real number with the following properties: (a) If > 0; there is an a1 in Œa; x0/ such that limx x0 ! D f .x/ exists and is f .x/ > ˛ if a1 x < x0: (b) If > 0 and a1 is in Œa; x0/; then f .x/ < ˛ C for some x Œa1; x0/: 2 2.1 Exercises 1. Each of the following conditions fails to deﬁne a function on any domain. State why. (a) sin f .x/ x2 (c) 1 If C 2. C x D Œf .x/2 0 D f .x/ .x 3/.x 1 x C D r g , Dfg , and Df =g . ﬁnd Df , Df ˙ (b) ef .x/ D j (d) f .x/Œf .x/ 2/ and g.x/ x2 x D x j x2 1 D 16 7 px2 9; 3. Find Df . (a) f .x/ tan x D (c) f .x/ 1 D x.x 1/ (e) eŒf .x/2 x; f .x/ 0 D Section 2.1 FunctionsandLimits 49 (b) f .x/ (d) f .x/ D D 1 sin x 1 j j p sin x x 4. Find limx x0 f .x/, and justify your answers with an –ı proof. 2x C 1; x0 1 D ! C 1 (a) x2 (c) (e) x2 .x x0 0 D ; 1 x3 1 1/.x 2/ C x; x0 1 D (b) x3 x 8 2 (d) px; ; x0 x0 2 D 4 D 5. Prove that Deﬁnition 2.1.2 is unchanged if Eqn. (2.1.4) is replaced by f .x/ j L j < K; where K is any positive constant. (That is, limx tion 2.1.2 if and only if limx x0 f .x/ ! x0 f .x/ L according to Deﬁni- L according to the modiﬁed deﬁnition.) D 6. Use Theorem 2.1.4 and the known limits limx ! D x0 x ! D x0, limx x0 c ! D c to ﬁnd the indicated limits. x2 2x (a) lim 2 x ! C 2x3 C x 1 C 1 3 (b) lim x8 x4 1 1 (d) lim 1 x ! f .x/, if they exist. Use –ı proofs, where ap- x3 (c) lim x 1 ! Find limx ! plicable, to justify your answers. x2 2x f .x/ and limx C x0 ! x0 C 7. x j ; x0 0 D x (a) (c) C j x x j x2 x C 1 j ; 2 x0 D 1 (d) x2 C px (b) x cos sin x0 1 x sin j 2 D x0 0 D ; j 8. Prove: If h.x/ 0 for a < x < x0 and limx h.x/ exists, then limx 0. Conclude from this that if f2.x/ f1.x/ for a < x < x0, then x0 ! h.x/ x0 ! lim x0 ! x if both limits exist. f2.x/ f1.x/ lim x0 ! x 50 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 9. 10. 11. 12. 13. (a) Prove: If limx f .x/ x 0 < ! M if 0 < < x0 j x j .) j x j j x0 f .x/ exists, there is a constant M and a > 0 such that (We say then that f is bounded on < . x0 j ˚ ˇ ˇ (b) State similar results with “limx (c) State similar results with “limx Suppose that limx Ln (a) by using Theorem 2.1.4 and induction; (b) directly from Deﬁnition 2.1.2. HINT: You will ﬁnd Exercise 2.1.9 useful for .b/: x0 ” replaced by “limx x0 ” replaced by “limx L and n is a positive integer. Prove that limx x0 f .x/ .” .” D x0 x0 ! ! ! ! ! ! C x0 Œf .x/n D x0 f .x/ Prove: If limx Prove Theorem 2.1.6. (a) Using the hint stated after Theorem 2.1.6, prove that Theorem 2.1.3 remains L > 0, then limx f .x/ p D D x0 ! ! pL. valid with “limx x0 ” replaced by “limx ! (b) Repeat (a) for Theorem 2.1.4. f .x/ 14. Deﬁne the statement “limx 15. !1 L.” D .” x0 ! f .x/ if it exists, and justify your answer directly from Deﬁnition 2.1.7. Find limx 1 (a) x2 (d) e !1 1 C x sin x (b) sin x ˛ x j j (e) tan x .˛ > 0/ (c) .˛ 0/ sin x ˛ x j j x2 e2x (f ) e x0 ” replaced throughout by 16. Theorems 2.1.3 and 2.1.4 remain valid with “limx “limx ” (“limx !1 !1 ”). How would their proofs have to be changed? ! 17. Using the deﬁnition you gave in Exercise 2.1.14, show that x (a) lim x 1 !1 1 x2 1 D (b) lim x !1 2 1 j C 2 j x D 18. sin x does not exist (c) lim x !1 Find limx answers directly from the deﬁnition you gave in Exercise 2.1.14. !1 f .x/, if it exists, for each function in Exercise 2.1.15. Justify your 19. Deﬁne (a) lim x0 x ! 20. Find f .x/ D 1 (b) lim x0 x ! C f .x/ D 1 (c) lim x0 x ! C f .x/ D 1 (a) lim 0 x C ! (c) lim 0 x C ! 1 x3 1 x6 1 x0/2k .x C positive integer) (e) lim x x0 ! (k D (b) lim 0 x ! (d) lim 0 x ! 1 x3 1 x6 (f ) lim x0 x ! .x 1 x0/2k 1 C Section 2.1 FunctionsandLimits 51 21. Deﬁne (a) lim x0 x ! f .x/ D 1 (b) lim x0 x ! f .x/ D 1 22. Find (a) lim 0 x ! 1 x3 1 x0/2k .x positive integer) (c) lim x0 x ! (k D 23. Deﬁne (a) lim x !1 f .x/ D 1 24. Find (a) lim x !1 x2k (c) lim x !1 x2k 1 C (k=positive integer) (e) lim x !1 px sin x (b) lim 0 x ! 1 x6 (d) lim x0 x ! .x 1 x0/2k 1 C (b) lim x !1 f .x/ D 1 (b) lim x !1 x2k (d) lim x !1 x2k 1 C ex (f ) lim x !1 / and .c; f .x/ f .g.x// 1 !1 D D 25. Suppose that f and g are deﬁned on .a; g.x/ > a if x > c. Suppose also that limx and limx . Show that limx g.x/ / respectively, and that , 1 1 1 L, where L. L 26. (a) Prove: limx !1 D 1 x0 f .x/ does not exist (ﬁnite) if for some 0 > 0, every deleted !1 neighborhood of x0 contains points x1 and x2 such that ! (b) Give analogous conditions for the nonexistence of f .x1/ j f .x2/ j 0: f .x/; lim x0 ! x C f .x/; lim x0 ! x lim x !1 f .x/; and f .x/: lim !1 x 27. < x0 < Prove: If only if limx equal, in which case all three are equal. 1 f .x/ and limx 1 x0 ! , then limx x0 ! C x0 f .x/ exists in the extended reals if and ! f .x/ both exist in the extended reals and are In Exercises 2.1.28–2.1.30 consider only the case where at least one of L1 and L2 is . ˙1 28. Prove: If limx nate, then x0 f .x/ ! D L1, limx x0 g.x/ ! D L2, and L1 C L2 is not indetermi- .f lim x0 x ! C g/.x/ L1 D C L2: 52 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 29. Prove: If limx then !1 f .x/ L1, limx g.x/ D !1 D L2, and L1L2 is not indeterminate, lim x !1 .fg/.x/ L1L2: D 30. (a) Prove: If limx x0 f .x/ indeter
minate, then ! L1, limx D ! x0 g.x/ 0, and L1=L2 is not L2 ¤ : lim x0 x ! f g .x/ D L1 L2 D ¤ (b) Show that it is necessary to assume that L2 sin x, g.x/ cos x, and x0 =2. D D 0 in(a) by considering f .x/ D 31. Find (a) lim 0 x C ! x3 2x4 C C (c) lim x !1 2x4 x3 C C 3x2 2x 2x 3x2 3 2 C C 2 C 3 C (e) limx .ex2 ex/ !1 32. Find limx r .x/ and limx !1 !1 r .x/ D where an 0 and bm 0. ¤ ¤ (b) lim 0 x ! x3 2x4 C C 2x 3x2 3 2 C C !1 2x4 x3 3x2 2x C C (d) lim x 2 C 3 C px sin x (f ) lim x e x !1 r .x/ for the rational function C 2x C x a0 b0 a1x b1x C C C C C C anxn bmxm ; Suppose that limx D on a set S with no limit points in .a; b/. What can be said about limx x0 in .a; b/? Justify your answer. x0 f .x/ exists for every x0 in .a; b/ and g.x/ ! f .x/ except x0 g.x/ for ! Prove Theorem 2.1.9(b), and complete the proof of Theorem 2.1.9(b) in the case where f is nonincreasing. Prove Theorem 2.1.12. Show that if f is bounded on Œa; x0/, then (a) (b) (c) x lim x0 ! lim x0 x ! lim x x0 case ! f .x/ x lim x0 ! f .x/. f /.x/ . D x f .x/ D x lim x0 ! f .x/ f .x/ and lim x0 lim x0 ! f .x/ if and only if limx . ! x f /.x/ D x lim x0 f .x/. ! f .x/ exists, in which x0 ! lim x0 ! x lim x0 D x f .x/ D x lim x0 ! f .x/: Suppose that f and g are bounded on Œa; x0/. ! 33. 34. 35. 36. 37. Section 2.2 Continuity 53 (a) Show that (b) Show that lim x0 ! x .f C g/.x/ x lim x0 ! f .x/ C x lim x0 ! g.x/: lim x0 ! x .f C g/.x/ f .x/ x lim x0 ! C x lim x0 ! g.x/: (c) State inequalities analogous to those in (a) and (b) for lim x0 ! x .f g/.x/ and lim x0 ! x .f g/.x/: Prove: limx f .x2/ such that j show that f is bounded on some interval .a; x0/ and f .x/ exists (ﬁnite) if and only if for each > 0 there is a ı > 0 ı < x1, x2 < x0. HINT: For sufﬁciency; x0 ! f .x1/ < if x0 j f .x/ lim 0 ! x D x lim x0 ! f .x/: Then use Exercise 2.1.36.c/: x0 Suppose that f is bounded on an interval .x0; b. Using Deﬁnition 2.1.10 as a guide, f .x/ (the f .x/ (the right limit superior of f at x0) and limx deﬁne limx right limit inferior of f at x0). Then prove that they exist. HINT: Use Theorem 2.1.9: Suppose that f is bounded on an interval .x0; b. Show that limx limx f .x/ exists, in which case f .x/ if and only if limx f .x/ D x0 x0 ! ! ! C C C x0 ! C x0 ! C 38. 39. 40. f .x/ lim x0 ! x C D x lim x0 ! C f .x/ D x lim x0 ! C f .x/: 41. Suppose that f is bounded on an open interval containing x0. Show that limx exists if and only if x0 f .x/ ! f .x/ lim x0 ! x D x lim x0 ! C f .x/ D x lim x0 ! f .x/ D x lim x0 ! C f .x/; in which case limx x0 f .x/ is the common value of these four expressions. ! 2.2 CONTINUITY In this section we study continuous functions of a real variable. We will prove some important theorems about continuous functions that, although intuitively plausible, are beyond the scope of the elementary calculus course. They are accessible now because of our better understanding of the real number system, especially of those properties that stem from the completeness axiom. 54 Chapter 2 DifferentialCalculusofFunctionsofOneVariable The deﬁnitions of f .x0 / D x lim x0 ! f .x/; f .x0 / C D x lim x0 ! C f .x/; and f .x/ lim x0 x ! do not involve f .x0/ or even require that it be deﬁned. However, the case where f .x0/ is deﬁned and equal to one or more of these quantities is important. Deﬁnition 2.2.1 (a) We say that f is continuous at x0 if f is deﬁned on an open interval .a; b/ containing x0 and limx x0 f .x/ f .x0/. ! D (b) We say that f is continuous from the left at x0 if f is deﬁned on an open interval .a; x0/ and f .x0 / D f .x0/. (c) We say that f is continuous from the right at x0 if f is deﬁned on an open interval .x0; b/ and f .x0 / C D f .x0/. The following theorem provides a method for determining whether these deﬁnitions are satisﬁed. The proof, which we leave to you (Exercise 2.2.1), rests on Deﬁnitions 2.1.2, 2.1.5, and 2.2.1. Theorem 2.2.2 (a) A function f is continuous at x0 if and only if f is deﬁned on an open interval .a; b/ containing x0 and for each > 0 there is a ı > 0 such that f .x/ j f .x0/ j < (2.2.1) < ı: whenever x x0 j j C (b) A function f is continuous from the right at x0 if and only if f is deﬁned on an interval Œx0; b/ and for each > 0 there is a ı > 0 such that (2.2.1) holds whenever x0 x < x0 ı: (c) A function f is continuous from the left at x0 if and only if f is deﬁned on an interval .a; x0 and for each > 0 there is a ı > 0 such that (2.2.1) holds whenever x0 ı < x x0: From Deﬁnition 2.2.1 and Theorem 2.2.2, f is continuous at x0 if and only if or, equivalently, if and only if it is continuous from the right and left at x0 (Exercise 2.2.2). f .x0 / D f .x0 / C D f .x0/ Example 2.2.1 Let f be deﬁned on Œ0; 2 by f .x/ x2; x C 0 1; 1 x < 1; 2 x D (Figure 2.2.1); then Section 2.2 Continuity 55 2; D f .0 f .1 f .1 f .0/; f .1/ f .1/; f .2/: D D Therefore, f is continuous from the right at 0 and 1 and continuous from the left at 2, but not at 1. If 0 < x, x0 < 1, then D D f .x/ f .x0/ x2 j j D j x 2 j Hence, f is continuous at each x0 in .0; 1/. If 1 < x, x0 < 2, then x2 x 0j D j < x0 j x0 x0 C j j j if x x x0 j j < =2: f .x/ j f .x0/ .x j D j < 1/ C if x j .x0 C x0 j x 1/ < : D j x0 j Hence, f is continous at each x0 in .1; 2/. y 3 2 1 y = x + 1, 0 ≤ x < 1 1 2 x Figure 2.2.1 Deﬁnition 2.2.3 A function f is continuous on an open interval .a; b/ if it is continuous at every point in .a; b/. If, in addition, or f .b / D f .b/ f .a / C D f .a/ (2.2.2) (2.2.3) 56 Chapter 2 DifferentialCalculusofFunctionsofOneVariable then f is continuous on .a; b or Œa; b/, respectively. If f is continuous on .a; b/ and (2.2.2) and (2.2.3) both hold, then f is continuous on Œa; b. More generally, if S is a subset of Df consisting of ﬁnitely or inﬁnitely many disjoint intervals, then f is continuous on S if f is continuous on every interval in S . (Henceforth, in connection with functions of one variable, whenever we say “f is continuous on S ” we mean that S is a set of this kind.) Example 2.2.2 Let f .x/ px, 0 f .x/ j D f .0/ j D x < 1 px < so f .0 / C D f .0/. If x0 > 0 and x 0, then . Then if 0 x < 2; f .x/ j f .x0/ j D j px0 x j px j D j < if px x0 x px0 j C x j x0 j px0 x0 j /. 1 < px0; so limx x0 f .x/ ! D f .x0/. Hence, f is continuous on Œ0; Example 2.2.3 The function g.x/ 1 sin x D is continuous on S (integer), since it is not deﬁned at such points. .n; n D1 C D 1n 1/. However, g is not continuous at any x0 n D S The function f deﬁned in Example 2.2.1 (see also Figure 2.2.1) is continuous on Œ0; 1/ and Œ1; 2, but not on any open interval containing 1. The discontinuity of f there is of the simplest kind, described in the following deﬁnition. C Deﬁnition 2.2.4 A function f is piecewise continuous on Œa; b if (a) f .x0 (b) f .x0 (c) f .x0 If (c) fails to hold at some x0 in .a; b/, f has a jump discontinuity at x0. Also, f has a f .a/ or at b if f .b jump discontinuity at a if f .a / exists for all x0 in Œa; b/; / exists for all x0 in .a; b; / f .x0/ for all but ﬁnitely many points x0 in .a; b/. f .b/. f .x0 / D D C / C ¤ / ¤ Example 2.2.4 The function f .x/ D 0; 1; x D x; 0 < x < 1; 2; x 1; x; 1 < x 2; 1; 2 < x < 3; 0; x D 3; D 8 ˆˆˆˆˆˆ< ˆˆˆˆˆˆ: (Figure 2.2.2) is the graph of a piecewise continuous function on Œ0; 3, with jump discontinuities at x0 0, 1, 2, and 3. D Section 2.2 Continuity 57 y 3 2 1 −1 1 2 3 x Figure 2.2.2 The reason for the adjective “jump” can be seen in Figures 2.2.1 and 2.2.2, where the graphs exhibit a deﬁnite jump at each point of discontinuity. The next example shows that not all discontinuities are of this kind. Example 2.2.5 The function f .x/ sin D 8 ˆ< 0; 1 x ; x x 0; 0; ¤ D ˆ: 1 and 1 with ever-increasing frequency, so neither f .0 0. As x approaches 0 from either side, f .x/ oscillates is continuous at all x0 except x0 / exists. between Therefore, the discontinuity of f at 0 is not a jump discontinuity, and if > 0, then f is not piecewise continuous on any interval of the form Œ ; , or Œ0; . / nor f .0 ; 0, Œ D C Theorems 2.1.4 and 2.2.2 imply the next theorem (Exercise 2.2.18). Theorem 2.2.5 If f and g are continuous on a set S; then so are f fg: In addition; f =g is continuous at each x0 in S such that g.x0/ 0: ¤ g; f C g; and Example 2.2.6 Since the constant functions and the function f .x/ x are continuous for all x, successive applications of the various parts of Theorem 2.2.5 imply that the function D r .x/ D 9 x x2 1 C 58 Chapter 2 DifferentialCalculusofFunctionsofOneVariable is continuous for all x except x from Theorem 2.2.5 and using 1 (see Example 2.1.7). More generally, by starting D induction, it can be shown that if f1, f2, . . . , fn are continuous on a set S , then so are f1 f2 C C C fn and f1f2 r .x/ D fn. Therefore, any rational function anxn bmxm .bm a1x b1x 0/ ¤ a0 b0 C C C C C C is continuous for all values of x except those for which its denominator vanishes. Removable Discontinuities Let f be deﬁned on a deleted neighborhood of x0 and discontinuous (perhaps even undeﬁned) at x0. We say that f has a at x0 if limx x0 f .x/ exists. In this case, the function ! if x f .x/ if x Df and x x0; ¤ x0; 2 D g.x/ f .x/ lim D 8 x0 x < ! : is continuous at x0. Example 2.2.7 The function f .x/ x sin D 1 x is not deﬁned at x0 (Example 2.1.6). Therefore, f has a removable discontinuity at 0. 0, and therefore certainly not continuous there, but limx D 0 f .x/ ! 0 D The function f1.x/ sin D 1 x is undeﬁned at 0 and its discontinuity there is not removable, since limx exist (Example 2.2.5). 0 f1.x/ does not ! Composite Functions We have seen that the investigation of limits and continuity can be simpliﬁed by regarding a given function as the result of addition, subtraction, multiplication, and division of simpler functions. Another operation useful in this connection is composition of functions; that is, substitution of one function into another. Deﬁnition 2.2.6 Suppose that f and g are functions with domains Df and Dg . If T , then the composite Dg has a nonempty subset T such that g.x/ function f g is
deﬁned on T by Df whenever x 2 2 ı g/.x/ .f ı D f .g.x//: f .x/ D log x and g.x/ Example 2.2.8 If then Since g.x/ > 0 if x 2 .0; / 1 Df T D . D and Dg D 1; 1/, the composite function f ˚ 1 ˇ ˇ x2 : ı f is deﬁned on .0; 1=e/ g/.x/ log D 1 f /.x/ 1 D 1 .log x/2 : .f We leave it to you to verify that g .g ı ı Section 2.2 Continuity 59 1 ; x2 ¤ ˙ 1 : g is deﬁned on . 1; 1/ by ı D 1 x x .1=e; e/ .e; [ / by 1 [ The next theorem says that the composition of continuous functions is continuous. Theorem 2.2.7 Suppose that g is continuous at x0; g.x0/ is an interior point of Df ; and f is continuous at g.x0/: Then f g is continuous at x0: ı Proof Suppose that > 0. Since g.x0/ is an interior point of Df and f is continuous at g.x0/, there is a ı1 > 0 such that f .t/ is deﬁned and f .t/ j f .g.x0// j < if t j g.x0/ j < ı1: Since g is continuous at x0, there is a ı > 0 such that g.x/ is deﬁned and Now (2.2.4) and (2.2.5) imply that g.x/ j g.x0/ j < ı1 if x j x0 j < ı: (2.2.4) (2.2.5) f .g.x// j f .g.x0// j < if x j x0 j < ı: g is continuous at x0. Therefore, f ı See Exercise 2.2.22 for a related result concerning limits. Example 2.2.9 In Examples 2.2.2 and 2.2.6 we saw that the function is continuous for x > 0, and the function f .x/ px D g.x/ 9 x D x2 1 C is continuous for x ¤ implies that the function 1. Since g.x/ > 0 if x < 3 or 1 < x < 3, Theorem 2.2.7 is continuous on . ; 1 3/ [ . 9 x x2 1 ı .f g/.x/ C 1; 3/. It is also continuous from the left at D s 3 and 3. 60 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Bounded Functions A function f is bounded below on a set S if there is a real number m such that In this case, the set f .x/ m for all x S: 2 V D f .x/ x S 2 has an inﬁmum ˛, and we write ˚ D If there is a point x1 in S such that f .x1/ and write ˛ ˇ ˇ f .x/: inf S x 2 ˛, we say that ˛ is the minimum of f on S , D D Similarly, f is bounded above on S if there is a real number M such that f .x/ all x in S . In this case, V has a supremum ˇ, and we write ˛ f .x/: min S x 2 M for f .x/: ˇ D sup S x 2 If there is a point x2 in S such that f .x2/ and write D ˇ, we say that ˇ is the maximum of f on S , If f is bounded above and below on a set S , we say that f is bounded on S . ˇ D max S x 2 f .x/: Figure 2.2.3 illustrates the geometric meaning of these deﬁnitions for a function f Œa; b. The graph of f lies in the strip bounded by the bounded on an interval S lines y m, where M is any upper bound and m is any lower bound for f on Œa; b. The narrowest strip containing the graph is the one bounded above by ˛ y b f .x/ and below by y M and y b f .x/. supa infa Figure 2.2.3 Example 2.2.10 The function g.x/ D ( 1 2 ; 1 x 0 or D x; 0 < x < 1; Section 2.2 Continuity 61 1; x D C (Figure 2.2.4(a)) is bounded on Œ0; 1, and sup x 0 g.x/ 1; D 1 inf x 0 1 g.x/ 0: D Therefore, g has no maximum or minimum on Œ0; 1, since it does not assume either of the values 0 and 1. The function which differs from g only at 0 and 1 (Figure 2.2.4(b)), has the same supremum and inﬁmum as g, but it attains these values at x 1, respectively; therefore, 0 and x h.x/ 1 x; 0 x 1; D D 1 and D min x 1 0 h.x/ 0: D y 1 1 2 max 1 x 0 h.x/ D y = g(x) 1 (ab) x Example 2.2.11 The function Figure 2.2.4 f .x/ D ex.x 1/ sin 1 ; 1/ x.x 0 < x < 1; oscillates between ˙ where 0 < < 1, and ex.x 1/ inﬁnitely often in every interval of the form .0; / or .1 ; 1/, sup 0<x<1 f .x/ 1; D inf 0<x<1 f .x/ 1: D However, f does not assume these values, so f has no maximum or minimum on .0; 1/. D f .ti / j C if M D x 1 62 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Theorem 2.2.8 If f is continuous on a ﬁnite closed interval Œa; b; then f is bounded on Œa; b: Proof Suppose that t containing t such that 2 Œa; b. Since f is continuous at t, there is an open interval It f .x/ j f .t/ j < 1 if x It 2 \ Œa; b: (2.2.6) 1 in (2.2.1), Theorem 2.2.2.) The collection H (To see this, set is an open covering of Œa; b. Since Œa; b is compact, the Heine–Borel theorem implies that there are ﬁnitely many points t1, t2, . . . , tn such that the intervals It1, It2, . . . , Itn cover Œa; b. According to (2.2.6) with t D D It a ˇ ˇ b ˚ t ti , f .x/ j < 1 if x Iti \ 2 Œa; b: (2.2.7) Œa; b. Therefore, j Let f .x/ .f .x/ f .ti // f .ti / j D j 1 f .ti / j C j j j Iti \ 2 Œa; b: f .x/ f .ti / j C j f .ti / j C max i 1 n j f .ti / j : Since Œa; b n i 1 D Iti \ Œa; b , (2.2.7) implies that f .x/ M if x j j 2 This proof illustrates the utility of the Heine–Borel theorem, which allows us to choose M as the largest of a ﬁnite set of numbers. S Theorem 2.2.8 and the completeness of the reals imply that if f is continuous on a ﬁnite closed interval Œa; b, then f has an inﬁmum and a supremum on Œa; b. The next theorem shows that f actually assumes these values at some points in Œa; b. Theorem 2.2.9 Suppose that f is continuous on a ﬁnite closed interval Œa; b: Let ˛ D a inf x f .x/ and ˇ b D a sup x f .x/: b Then ˛ and ˇ are respectively the minimum and maximum of f on Œa; b points x1 and x2 in Œa; b such that I that is; there are f .x1/ D ˛ and f .x2/ ˇ: D Proof We show that x1 exists and leave it to you to show that x2 exists (Exercise 2.2.24). ˛. Then f .x/ > ˛ for all Suppose that there is no x1 in Œa; b such that f .x1/ Œa; b. We will show that this leads to a contradiction. D x 2 Suppose that t Œa; b. Then f .t/ > ˛, so 2 f .t/ > ˛ f .t/ C 2 > ˛: Section 2.2 Continuity 63 Since f is continuous at t, there is an open interval It about t such that f .x/ > ˛ f .t/ C 2 if x It 2 \ Œa; b (2.2.8) (Exercise 2.2.15). The collection H is an open covering of Œa; b. Since a Œa; b is compact, the Heine–Borel theorem implies that there are ﬁnitely many points t1, t2, . . . , tn such that the intervals It1, It2 , . . . , Itn cover Œa; b. Deﬁne D It ˇ ˇ b ˚ t Then, since Œa; b n i 1.Iti \ D ˛1 D min n i 1 f .ti / 2 ˛ : C Œa; b/, (2.2.8) implies that S f .t/ > ˛1; a t b: But ˛1 > ˛, so this contradicts the deﬁnition of ˛. Therefore, f .x1/ Œa; b. D ˛ for some x1 in Example 2.2.12 We used the compactness of Œa; b in the proof of Theorem 2.2.9 when we invoked the Heine–Borel theorem. To see that compactness is essential to the proof, consider the function g.x/ .1 1 D x/ sin 1 x ; which is continuous and has supremum 2 on the noncompact interval .0; 1, but does not assume its supremum on .0; 1, since g.x/ .1 .1 1 1 C C 1 x x/ sin ˇ ˇ x/ < 2 if : As another example, consider the function D which is continuous and has inﬁmum 0, which it does not attain, on the noncompact interval .0; /. 1 f .x/ e x; The next theorem shows that if f is continuous on a ﬁnite closed interval Œa; b, then f assumes every value between f .a/ and f .b/ as x varies from a to b (Figure 2.2.5, page 64). Theorem 2.2.10 (Intermediate Value Theorem) Suppose that f is con for tinuous on Œa; b; f .a/ some c in .a; b/: f .b/; and is between f .a/ and f .b/: Then f .c/ ¤ D 64 Chapter 2 DifferentialCalculusofFunctionsofOneVariable y y = f (x) y = µ x a x b Figure 2.2.5 Proof Suppose that f .a/ < < f .b/. The set D S D x a x b and f .x/ sup S . We will show that f .c/ ˇ D ˇ < x . If f .c/ > , is bounded and nonempty. Let c ˚ then c > a and, since f is continuous at c, there is an > 0 such that f .x/ > if c is an upper bound for S , which contradicts the deﬁnition of c as the supremum of S . If f .c/ < , then c < b and there is , so c is not an upper bound for S . This is an > 0 such that f .x/ < for c also a contradiction. Therefore, f .c/ c (Exercise 2.2.15). Therefore, c x < c . C D The proof for the case where f .b/ < < f .a/ can be obtained by applying this result to f . Uniform Continuity Theorem 2.2.2 and Deﬁnition 2.2.3 imply that a function f is continuous on a subset S of its domain if for each > 0 and each x0 in S , there is a ı > 0, which may depend upon x0 as well as , such that f .x/ j f .x0/ j < if x j x0 j < ı and x Df : 2 The next deﬁnition introduces another kind of continuity on a set S . Deﬁnition 2.2.11 A function f is uniformly continuous on a subset S of its domain if, for every > 0, there is a ı > 0 such that f .x/ f .x0/ j < whenever x x0j < ı and x; x0 2 S: j We emphasize that in this deﬁnition ı depends only on and S and not on the particular j choice of x and x0, provided that they are both in S . Example 2.2.13 The function f .x/ 2x D is uniformly continuous on . /, since ; 1 f .x0/ 1 2 f .x/ j Example 2.2.14 If 0 < r < j D j x if < x0j , then the function 1 is uniformly continuous on Œ r; r . To see this, note that g.x/ x2 D g.x/ j g.x0/ D j x2 .x0/2 x j D j so g.x/ j g.x0/ j < if x j x0j < ı x x0j j 2r D Section 2.2 Continuity 65 x j x0j < =2: x0j 2r x j ; x0j C and r x; x0 r: Often a concept is clariﬁed by considering its negation: a function f is not uniformly continuous on S if there is an 0 > 0 such that if ı is any positive number, there are points x and x0 in S such that x j x0j < ı but Example 2.2.15 The function g.x/ ﬁnite r (Example 2.2.14), but not on . there are real numbers x and x0 such that D ; 1 f .x/ f .x0/ 0: j j x2 is uniformly continuous on Œ r; r for any /. To see this, we will show that if ı > 0 1 x j x0j D ı=2 and g.x/ j g.x0/ j 1: To this end, we write g.x/ g.x0/ j j D j ı=2 and x; x0 > 1=ı, then x2 .x0/2 x x0 x j j x0 : j C j D j If x j x0j D x j x0 x j j x0 : Example 2.2.16 The function f .x/ cos D 1 x is continuous on .0; 1 (Exercise 2.2.23(i)). However, f is not uniformly continuous on .0; 1, since 1 n f 1 .n C 1/ 2; n D D 1; 2 Examples 2.2.15 and 2.2.16 show that a function may be continuous but not uniformly continuous on an interval. The next theorem shows that this cannot happen if the interval is closed and bounded, and therefore compact. 66 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Theorem 2.2.12 If f is continuous on a closed and bounded interval Œa; b; then f is uniformly continuous on Œa; b: Proof Suppose that > 0. Since f is continuous on Œa; b, for each t in Œa; b there is a positive number ıt such that f .x/ if < f .t/ j ıt /, the collection j ıt ; t 2 If It .t D
C x j t j < 2ıt and x Œa; b: 2 (2.2.9) H It t 2 D Œa; b ˇ is an open covering of Œa; b. Since Œa; b is compact, the Heine–Borel theorem implies that ˇ there are ﬁnitely many points t1, t2, . . . , tn in Œa; b such that It1, It2 , . . . , Itn cover Œa; b. Now deﬁne ˚ We will show that if ı min f ıt1; ıt2; : : : ; ıtn g : D x j < ı and x0j x; x0 2 Œa; b; then f .x/ f .x0/ j From the triangle inequality, < . j f .x/ j f .x0/ j D j j .f .x/ f .x/ f .tr // f .tr / .f .tr / f .tr / f .x0// : f .x0/ j C j C j (2.2.10) (2.2.11) j (2.2.12) Since It1 , It2, . . . , Itn cover Œa; b, x must be in one of these intervals. Suppose that x that is, 2 Itr ; x j tr j < ıtr : From (2.2.9) with t tr , D f .tr / j From (2.2.11), (2.2.13), and the triangle inquality, f .x/ j < 2 : tr .x0 x0 Therefore, (2.2.9) with t j D j j x/ C .x tr / x0 j tr and x replaced by x0 implies that j j j C j tr x x D (2.2.13) (2.2.14) < ı ıtr C 2ıtr : f .x0/ j 2 : < f .tr / j f .x0/ j This, (2.2.12), and (2.2.14) imply that f .x/ < . This proof again shows the utility of the Heine–Borel theorem, which allowed us to deﬁne ı in (2.2.10) as the smallest of a ﬁnite set of positive numbers, so that ı is sure to be positive. (An inﬁnite set of positive numbers may fail to have a smallest positive member; for example, consider the open interval .0; 1/.) j Corollary 2.2.13 If f is continuous on a set T; then f is uniformly continuous on any ﬁnite closed interval contained in T: Section 2.2 Continuity 67 Applied to Example 2.2.16, Corollary 2.2.13 implies that the function g.x/ is uniformly continuous on Œ; 1 if 0 < < 1. cos 1=x D More About Monotonic Functions Theorem 2.1.9 implies that if f is monotonic on an interval I , then f is either continuous or has a jump discontinuity at each x0 in I . This and Theorem 2.2.10 provide the key to the proof of the following theorem. Theorem 2.2.14 If f is monotonic and nonconstant on Œa; b; then f is continuous on Œa; b if and only if its range Rf is the closed interval with endpoints x f .a/ and f .b/: Œa; b f .x/ D 2 ˚ ˇ ˇ Proof We assume that f is nondecreasing, and leave the case where f is nonincreasing to you (Exercise 2.2.34). Theorem 2.1.9(a) implies that the set .a; b/ is a subset of the open interval .f .a //. Therefore, /; f .b f .x/ Rf D 2 x Rf f .a/ Rf f .b/ g [ D f [ f Now suppose that f is continuous on Œa; b. Then f .a/ e (2.2.15) implies that Rf implies that f .x/ for some x in .a; b/. Hence, Rf g f g [ C .f .a /; f .b f .b/, so D Œf .a/; f .b/. If f .a/ < < f .b/, then Theorem 2.2.10 /, f .b / D C ˇ ˇ ˚ f .b/ g : [ f (2.2.15) e // f .a C f .a/ D For the converse, suppose that Rf Œf .a/; f .b/. Since f .a/ f .b/, (2.2.15) implies that f .a/ rem 2.1.9(c) that if f is nondecreasing and a < x0 < b, then / and f .b / C D D D f .a D Œf .a/; f .b/. f .a f .b/. We know from Theo- C / and f .b / f .x0 / f .x0/ f .x0 /: C If either of these inequalities is strict, Rf cannot be an interval. Since this contradicts our assumption, f .x0 /. Therefore, f is continuous at x0 (Exerf .x0/ cise 2.2.2). We can now conclude that f is continuous on Œa; b. f .x0 / D D C Theorem 2.2.14 implies the following theorem. Theorem 2.2.15 Suppose that f is increasing and continuous on Œa; b; and let f .a/ c and f .b/ d: Then there is a unique function g deﬁned on Œc; d such that D D and g.f .x// x; a x b; D Moreover; g is continuous and increasing on Œc; d : D f .g.y// y; c d: y (2.2.16) (2.2.17) Proof We ﬁrst show that there is a function g satisfying (2.2.16) and (2.2.17). Since f is continuous, Theorem 2.2.14 implies that for each y0 in Œc; d there is an x0 in Œa; b such that f .x0/ y0; D (2.2.18) 68 Chapter 2 DifferentialCalculusofFunctionsofOneVariable and, since f is increasing, there is only one such x0. Deﬁne g.y0/ x0: D (2.2.19) The deﬁnition of x0 is illustrated in Figure 2.2.6: with Œc; d drawn on the y-axis, ﬁnd the f .x/ and drop a vertical from the intersection of the line y intersection to the x-axis to ﬁnd x0. y0 with the curve y D D y d y0 c y = f (x) a x0 b x Figure 2.2.6 Substituting (2.2.19) into (2.2.18) yields and substituting (2.2.18) into (2.2.19) yields f .g.y0// y0; D g.f .x0// x0: D Dropping the subscripts in these two equations yields (2.2.16) and (2.2.17). The uniqueness of g follows from our assumption that f is increasing, and therefore only one value of x0 can satisfy (2.2.18) for each y0. To see that g is increasing, suppose that y1 < y2 and let x1 and x2 be the points in Œa; b y1 and f .x2/ y2. Since f is increasing, x1 < x2. Therefore, such that f .x1/ D D g.y1/ D g.y/ x1 < x2 g.y2/; D Œc; d so g is increasing. Since Rg Œa; b, 2 Theorem 2.2.14 with f and Œa; b replaced by g and Œc; d implies that g is continuous on Œc; d . is the interval Œg.c/; g.d / D D y ˚ ˇ ˇ The function g of Theorem 2.2.15 is the inverse of f , denoted by f 1. Since (2.2.16) and (2.2.17) are symmetric in f and g, we can also regard f as the inverse of g, and denote it by g 1. then 1. 2. Example 2.2.17 If then Example 2.2.18 If Section 2.2 Continuity 69 f .x/ x2; D x 0 R; f 1.y/ g.y/ D D py; 0 y R2: f .x/ 2x D f 1.y/ g.y/ D 4; 0 x 2; y 4 ; 2 4 y 8: C D 2.2 Exercises Prove Theorem 2.2.2. Prove that a function f is continuous at x0 if and only if f .x/ lim x0 ! x D x lim x0 ! C f .x/ D f .x0/: 3. Determine whether f is continuous or discontinuous from the right or left at x0. (b) f .x/ (a) f .x/ 0/ (c) f .x/ (e) f .x/ .x0 D 0/ px .x0 1 x x sin 1=x; x x 1; D D D D ¤ D x sin 1=x; x 0; x x x ¤ D x/ C j .1 j x C (f ) f .x/ D (g) f .x/ D 8 < 1; 4. Let f be deﬁned on Œ0; 2 by : (d) f .x/ px .x0 > 0/ x2 .x0 arbitrary/ D D 0; 0 0 0 sin .x0 0/ D .x0 0x0 0/ D f .x/ x2; 0 D ( x C 1; 1 x < 1; 2: x On which of the following intervals is f continuous according to Deﬁnition 2.2.3: Œ0; 1/, .0; 1/, .0; 1, Œ0; 1, Œ1; 2/, .1; 2/, .1; 2, Œ1; 2? 5. Let g.x/ px 1 x : D On which of the following intervals is g continuous according to Deﬁnition 2.2.3: Œ0; 1/, .0; 1/, .0; 1, Œ1; /, .1; /? 1 1 70 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 6. Let f .x/ -1 if x is irrational; D ( 1 if x is rational: Show that f is not continuous anywhere. 7. Let f .x/ 0 if x is irrational and f .p=q/ 1=q if p and q are positive integers with no common factors. Show that f is discontinuous at every rational and continuous at every irrational on .0; D D /. 8. Prove: If f assumes only ﬁnitely many values, then f is continuous at a point x0 in D0 f if and only if f is constant on some interval .x0 ı; x0 ı/. 9. The characteristic function T of a set T is deﬁned by C 1 T .x/ 1; x D ( 0; x T; T: 2 62 10. 11. 12. 13. 14. 15. Show that T is continuous at a point x0 if and only if x0 Prove: If f and g are continuous on .a; b/ and f .x/ subset (Deﬁnition 1.1.5) of .a; b/, then f .x/ D g.x/ for all x in .a; b/. T 0 .T c/0. 2 [ g.x/ for every x in a dense D log x is continuous on .0; 0. D D D 1 1 1 g.x/ ; 1 D g.x2/ 0 f .x/ x2/ /. Take the following /. Take the following g.x1x2/ if x1; x2 > 0. eax is continuous on . 1. f .x1/f .x2/; Prove that the function g.x/ properties as given. (a) limx ! (b) g.x1/ C Prove that the function f .x/ properties as given. (a) limx ! (b) f .x1 (a) Prove that the functions sinh x and cosh x are continuous for all x. (b) For what values of x are tanh x and coth x continuous? Prove that the functions s.x/ Take the following properties as given. (a) limx (b) c.x1 (c) s2.x/ . 1 (a) Prove: If f is continuous at x0 and f .x0/ > , then f .x/ > for all x in 0 c.x/ x2/ D c2.x/ cos x are continuous on . sin x and c.x/ ! C s.x1/s.x2/; < x1; x2 < c.x1/c.x2/ < x1; x2 < ; ; 1. /. some neighborhood of x0. (b) State a result analogous to (a) for the case where f .x0/ < . (c) Prove: If f .x/ for all x in S and x0 is a limit point of S at which f is (d) State results analogous to (a), (b), and (c) for the case where f is contin- continuous, then f .x0/ . uous from the right or left at x0. Section 2.2 Continuity 71 16. Let be the function whose value at each x in Df is continuous at x0, then so is f . Is the converse true? f j j f .x/ j j . Prove: If f is j j Prove: If f is monotonic on Œa; b, then f is piecewise continuous on Œa; b if and only if f has only ﬁnitely many discontinuities in Œa; b. Prove Theorem 2.2.5. (a) Show that if f1, f2, . . . , fn are continuous on a set S then so are f1 f2 C C C fn and f1f2 fn. (b) Use (a) to show that a rational function is continuous for all values of x 17. 18. 19. except the zeros of its denominator. 20. (a) Let f1 and f2 be continuous at x0 and deﬁne D Show that F is continuous at x0. F .x/ max .f1.x/; f2.x// : (b) Let f1, f2, . . . , fn be continuous at x0 and deﬁne F .x/ D max .f1.x/; f2.x/; : : : ; fn.x// : Show that F is continuous at x0. 21. 22. D 1 Find the domains of f (a) f .x/ px; ı g.x/ g and g 1 D f . ı x2 (b) f .x/ g.x/ cos x (d) f .x/ (c) f .x/ x2 ; D (a) Suppose that y0 1 D limx D that f is continuous at y0. Show that ! log x; g.x/ sin x D px; g.x/ sin 2x D D D x0 g.x/ exists and is an interior point of Df , and .f lim x0 x ! ı g/.x/ D f .y0/: (b) State an analogous result for limits from the right. (c) State an analogous result for limits from the left. 23. Use Theorem 2.2.7 to ﬁnd all points x0 at which the following functions are contin- uous. (a) p1 x2 (d) e 1=.1 2x/ (b) sin e x2 (e) sin 1 .x 1/2 (c) log.1 sin x/ C (f ) sin 1 cos x 1 x (g) .1 sin2 x/ 1=2 (h) cot.1 x2 / e (i) cos 24. Complete the proof of Theorem 2.2.9 by showing that there is an x2 such that f .x2/ ˇ. D 72 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 25. 26. I y f .x/; x D is an interval. Moreover, if I is a ﬁnite closed interval, then Prove: If f is nonconstant and continuous on an interval I , then the set S y D so is S . ˇ ˚ ˇ Suppose that f and g are deﬁned on . tinuous on . /. Show that g is continuous on . /, f is increasing, and f ; 1 g is con- ; 1 1 1 2 /. ı ; 1 27. Let f be continuous on Œa; b/, and deﬁne 1 F .x/ D max x t a f .t/; a x < b: (How do we know that F is well deﬁned?) Show that F is continuous on Œa; b/.
28. Let f and g be uniformly continuous on an interval S . (a) Show that f g are uniformly continuous on S . (b) Show that fg is uniformly continuous on S if S is compact. (c) Show that f =g is uniformly continuous on S if S is compact and g has no g and f C zeros in S . (d) Give examples showing that the conclusion of (b) and (c) may fail to hold if S is not compact. (e) State additional conditions on f and g which guarantee that fg is uniformly continuous on S even if S is not compact. Do the same for f =g. 29. 30. 31. 32. 33. Suppose that f is uniformly continuous on a set S , g is uniformly continuous on a set T , and g.x/ g is uniformly continuous on T . S for every x in T . Show that f 2 ı (a) Prove: If f is uniformly continuous on disjoint closed intervals I1, I2, . . . , In, then f is uniformly continuous on n j D 1 Ij . (b) Is (a) valid without the word “closed”? S (a) Prove: If f is uniformly continuous on a bounded open interval .a; b/, then f .a / and f .b / exist and are ﬁnite. HINT: See Exercise 2.1.38: (b) Show that the conclusion in (a) does not follow if .a; b/ is unbounded. / and f . Prove: If f is continuous on Œa; continuous on Œa; / exists (ﬁnite), then f is uniformly 1 1 /. C Suppose that f is deﬁned on . / and has the following properties. x2/ D f .x1/f .x2/; < x1; x2 < : 1 1 C 1 ; 1 1 1 and (ii) f .x1 (i) lim 0 x ! f .x/ D Prove: (a) f .x/ > 0 for all x. (b) f .r x/ (c) If f .1/ Œf .x/r if r is rational. 1 then f is constant. D D Section 2.3 DifferentiableFunctionsofOneVariable 73 (d) If f .1/ D > 1, then f is increasing, f .x/ lim lim x !1 !1 eax has these properties if a > 0.) ; D 1 and x f .x/ 0: D (Thus, f .x/ D HINT: See Exercises 2.2.10 and 2.2.12: 34. Prove Theorem 2.2.14 in the case where f is nonincreasing. 2.3 DIFFERENTIABLE FUNCTIONS OF ONE VARIABLE In calculus you studied differentiation, emphasizing rules for calculating derivatives. Here we consider the theoretical properties of differentiable functions. In doing this, we assume that you know how to differentiate elementary functions such as xn, ex, and sin x, and we will use such functions in examples. Deﬁnition of the Derivative Deﬁnition 2.3.1 A function f is differentiable at an interior point x0 of its domain if the difference quotient f .x/ x f .x0/ x0 ; x0; x ¤ approaches a limit as x approaches x0, in which case the limit is called the derivative of f at x0, and is denoted by f 0.x0/; thus, f 0.x0/ lim x0 x ! D f .x/ x f .x0/ x0 : (2.3.1) It is sometimes convenient to let x f 0.x0/ D D x0 C lim 0 h ! h and write (2.3.1) as f .x0 C h/ h f .x0/ : If f is deﬁned on an open set S , we say that f is differentiable on S if f is differentiable at every point of S . If f is differentiable on S , then f 0 is a function on S . We say that f is continuously differentiable on S if f 0 is continuous on S . If f is differentiable on a neighborhood of x0, it is reasonable to ask if f 0 is differentiable at x0. If so, we denote the derivative of f 0 at x0 by f 00.x0/. This is the second derivative of f at x0, and it is also 1/ is deﬁned on a neighborhood of denoted by f .2/.x0/. Continuing inductively, if f .n 1/ at x0, then the nth derivative of f at x0, denoted by f .n/.x0/, is the derivative of f .n x0. For convenience we deﬁne the zeroth derivative of f to be f itself; thus D We assume that you are familiar with the other standard notations for derivatives; for f .0/ f: example, f .2/ D f 00; f .3/ f 000; D 74 Chapter 2 DifferentialCalculusofFunctionsofOneVariable and so on, and d nf dxn D f .n/: Example 2.3.1 If n is a positive integer and f .x/ xn; D f .x/ x f .x0/ x0 D xn x xn 0 x0 D x x x0 x0 then so k xn 1xk 0 ; n 1 0 Xk D f 0.x0/ lim x0 x ! D k xn 1xk 0 D nxn 0 1 : n 1 0 Xk D Since this holds for every x0, we drop the subscript and write f 0.x/ D nxn 1 or d dx .xn/ D 1: nxn To derive differentiation formulas for elementary functions such as sin x, cos x, and ex directly from Deﬁnition 2.3.1 requires estimates based on the properties of these functions. Since this is done in calculus, we will not repeat it here. Interpretations of the Derivative If f .x/ is the position of a particle at time x x0, the difference quotient ¤ f .x0/ x0 f .x/ x is the average velocity of the particle between times x0 and x. As x approaches x0, the average applies to shorter and shorter intervals. Therefore, it makes sense to regard the limit (2.3.1), if it exists, as the particle’s instantaneous velocity at time x0. This interpretation may be useful even if x is not time, so we often regard f 0.x0/ as the instantaneous rate of change of f .x/ at x0, regardless of the speciﬁc nature of the variable x. The derivative also has a geometric interpretation. The equation of the line through two points .x0; f .x0// and .x1; f .x1// on the curve y f .x/ (Figure 2.3.1) is D f .x0/ y D f .x1/ x1 C f .x0/ x0 .x x0/: Varying x1 generates lines through .x0; f .x0// that rotate into the line f .x0/ y D C f 0.x0/.x x0/ (2.3.2) Section 2.3 DifferentiableFunctionsofOneVariable 75 as x1 approaches x0. This is the tangent to the curve y Figure 2.3.2 depicts the situation for various values of x1. D f .x/ at the point .x0; f .x0//. y y = f (x) x0 x1 x Figure 2.3.1 y = f (x) y Tangent line x0 '' x1 x1 x1 x Figure 2.3.2 Here is a less intuitive deﬁnition of the tangent line: If the function approximates f so well near x0 that T .x/ f .x0/ m.x x0/ C D lim x0 x ! f .x/ x T .x/ x0 0; D we say that the line y D T .x/ is tangent to the curve y f .x/ at .x0; f .x0//. D 76 Chapter 2 DifferentialCalculusofFunctionsofOneVariable This tangent line exists if and only if f 0.x0/ exists, in which case m is uniquely determined f 0.x0/ (Exercise 2.3.1). Thus, (2.3.2) is the equation of the tangent line. by m D We will use the following lemma to study differentiable functions. Lemma 2.3.2 If f is differentiable at x0; then f .x/ D f .x0/ C Œf 0.x0/ C E.x/.x x0/; (2.3.3) where E is deﬁned on a neighborhood of x0 and Proof Deﬁne E.x/ f .x/ x D 8 < 0; lim x0 x ! E.x/ E.x0/ 0: D D f .x0/ x0 f 0.x0/; x x Df and x x0; ¤ x0: 2 D (2.3.4) Solving (2.3.4) for f .x/ yields (2.3.3) if x tion 2.3.1 implies that limx at x0. x0 E.x/ D : ! x0, and (2.3.3) is obvious if x x0. DeﬁniD 0 to make E continuous ¤ 0. We deﬁned E.x0/ D Since the right side of (2.3.3) is continuous at x0, so is the left. This yields the following theorem. Theorem 2.3.3 If f is differentiable at x0; then f is continuous at x0: The converse of this theorem is false, since a function may be continuous at a point without being differentiable at the point. Example 2.3.2 The function can be written as or as From (2.3.5), and from (2.3.6), f .x/ x j D j f .x/ D x; x > 0; f .x/ x; D x < 0: f 0.x/ D 1; x > 0; f 0.x/ 1; D x < 0: (2.3.5) (2.3.6) Neither (2.3.5) nor (2.3.6) holds throughout any neighborhood of 0, so neither can be used alone to calculate f 0.0/. In fact, since the one-sided limits and lim 0 C ! x lim 0 ! x f .x/ x f .x/ x f .0/ 0 f .0/ 0 D x lim 0 C ! x x D x lim 0 ! x 1 x D (2.3.7) (2.3.8) Section 2.3 DifferentiableFunctionsofOneVariable 77 are different, f .x/ x does not exist (Theorem 2.1.6); thus, f is not differentiable at 0, even though it is continuous at 0. f .0/ 0 lim 0 x ! Interchanging Diﬀerentiation and Arithmetic Operations The following theorem should be familiar from calculus. g; f C g; and fg; Theorem 2.3.4 If f and g are differentiable at x0; then so are f with (a) .f (b) .f (c) .fg/0 .x0/ f .x0/g0.x0/: The quotient f =g is differentiable at x0 if g.x0/ f 0.x0/ f 0.x0/ D f 0.x0/g.x0/ g0.x0/ I g.x0/ I g/0.x0/ g/0.x0/ 0; with C D C C D (d) f g 0 .x0/ D f 0.x0/g.x0/ ¤ f .x0/g0.x0/ : Œg.x0/2 Proof The proof is accomplished by forming the appropriate difference quotients and applying Deﬁnition 2.3.1 and Theorem 2.1.4. We will prove (c) and leave the rest to you (Exercises 2.3.9, 2.3.10, and 2.3.11). The trick is to add and subtract the right quantity in the numerator of the difference quotient for .fg/0 .x0/; thus, f .x/g.x/ x f .x0/g.x0/ x0 D D f .x/g.x/ f .x0/g.x/ x f .x0/g.x/ x0 f .x0/g.x0/ f .x/ x f .x0/ x0 g.x/ C g.x/ x g.x0/ x0 : C f .x0/ The difference quotients on the right approach f 0.x0/ and g0.x0/ as x approaches x0, and limx g.x0/ (Theorem 2.3.3). This proves (c). x0 g.x/ ! D The Chain Rule Here is the rule for differentiating a composite function. Theorem 2.3.5 (The Chain Rule) Suppose that g is differentiable at x0 and f is differentiable at g.x0/: Then the composite function h g; deﬁned by f D ı is differentiable at x0; with h.x/ D f .g.x//; h0.x0/ D f 0.g.x0//g0.x0/: 78 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Proof Since f is differentiable at g.x0/, Lemma 2.3.2 implies that f .t/ f .g.x0// D Œf 0.g.x0// E.t/Œt g.x0/; C where Letting t D g.x/ yields lim g.x0/ ! t E.t/ D E.g.x0// 0: D (2.3.9) f .g.x// f .g.x0// D Œf 0.g.x0// C E.g.x//Œg.x/ g.x0/: Since h.x/ D f .g.x//, this implies that h.x/ x h.x0/ x0 D Œf 0.g.x0// E.g.x// C g.x/ x g.x0/ x0 : (2.3.10) Since g is continuous at x0 (Theorem 2.3.3), (2.3.9) and Theorem 2.2.7 imply that lim x0 x ! Therefore, (2.3.10) implies that E.g.x// E.g.x0// 0: D D h0.x0/ lim x0 x ! D h.x/ x h.x0/ x0 D f 0.g.x0//g0 .x0/; as stated. Example 2.3.3 If then and f .x/ D sin x and g.x/ h.x/ D f .g.x// sin D 1 x ; 0; x ¤ 0; x ¤ D 1 x ; h0.x/ D f 0.g.x//g.x/ 1 x D cos 1 x2 ; x 0: ¤ It may seem reasonable to justify the chain rule by writing h.x/ x h.x0/ x0 f .g.x// x f .g.x// g.x/ D D f .g.x0// x0 f .g.x0// g.x0/ g.x/ x g.x0/ x0 f .g.x// g.x/ lim x0 x ! f .g.x0// g.x0/ D f 0.g.x0// and arguing that Section 2.3 DifferentiableFunctionsofOneVariable 79 (because limx x0 g.x/ ! D g.x0// and g.x/ x lim x0 x ! g.x0/ x0 D g0.x0/: However, this is not a valid proof (Exercise 2.3.13). One-Sided Derivatives One-sided limits of difference quotients such as (2.3.7) and (2.3.8) in Example 2.3.2 are called one-sided or right- and left-hand derivatives. That is, if f is deﬁned on Œx0; b/, the right-hand derivative of f at x0 is deﬁned to be .x0/ f 0 C D x lim x0 !
f .x/ x f .x0/ x0 C if the limit exists, while if f is deﬁned on .a; x0, the left-hand derivative of f at x0 is deﬁned to be f .x/ x f .x0/ x0 lim x0 ! if the limit exists. Theorem 2.1.6 implies that f is differentiable at x0 if and only if f 0 C and f 0 .x0/ exist and are equal, in which case f 0 .x0/ D x .x0/ .x0/ .x0/: f 0 f 0.x0/ D 1 and f 0 f 0 C .0/ D D 1. In Example 2.3.2, f 0 C .0/ D Example 2.3.4 If then x3; f .x/ D 8 < x2 sin 1 x 0; x ; x > 0; : 3x2; x < 0; (2.3.11) f 0.x/ 1 x Since neither formula in (2.3.11) holds for all x in any neighborhood of 0, we cannot simply differentiate either to obtain f 0.0/; instead, we calculate ; x > 0: D 8 < 2x sin cos 1 x : (2.3.12) .0/ f 0 C D x lim 0 C ! hence, f 0.0/ D f 0 C .0/ f 0 .0/ D x D f 0 lim 0 ! .0/ D 1 x 0 0 0 D x2 sin x x3 x 0. 0 D x lim 0 C ! x sin 1 x D 0; lim 0 x ! x2 0 I D 80 Chapter 2 DifferentialCalculusofFunctionsofOneVariable This example shows that there is a difference between a one-sided derivative and a onesided limit of a derivative, since f 0 f 0.x/ C does not exist. It also shows that a derivative may exist in a neighborhood of a point x0 ( 0 in this case), but be discontinuous at x0. Exercise 2.3.4 justiﬁes the method used in Example 2.3.4 to compute f 0.x/ for x 0, but, from (2.3.12), f 0.0 limx .0. ¤ Deﬁnition 2.3.6 (a) We say that f is differentiable on the closed interval Œa; b if f is differentiable on the open interval .a; b/ and f 0 C .a/ and f 0 .b/ both exist. (b) We say that f is continuously differentiable on Œa; b if f is differentiable on Œa; b, f 0.a f 0.b .a/ .b/ /. f 0 is continuous on .a; b/, f 0 C /, and f 0 C D D Extreme Values We say that f .x0/ is a local extreme value of f if there is a ı > 0 such that f .x/ does not change sign on f .x0/ More speciﬁcally, f .x0/ is a local maximum value of f if .x0 ı; x0 ı/ \ C Df : or a local minimum value of f if f .x/ f .x0/ (2.3.13) (2.3.14) for all x in the set (2.3.13). The point x0 is called a local extreme point of f , or, more speciﬁcally, a local maximum or local minimum point of f . f .x/ f .x0/ (2.3.15) y −1 −1 2 1 2 1 2 3 4 x Figure 2.3.3 Section 2.3 DifferentiableFunctionsofOneVariable 81 Example 2.3.5 If f .x/ D 1; x ; j j sin x 2 ; 1 p2 ˆˆˆˆ< ˆˆˆˆ: (Figure 2.3.3), then 0, 3, and every x in . 1; 4, and every x in . 1 2 are local maximum points. 1; 1 2 / are local minimum points of f , while 1, It is geometrically plausible that if the curve y f .x/ has a tangent at a local extreme point of f , then the tangent must be horizontal; that is, have zero slope. (For example, in Figure 2.3.3, see x 1=2/.) The following theorem shows that this must be so. 3, and every x in . 1, x D D D 1; Theorem 2.3.7 If f is differentiable at a local extreme point x0 D0 f ; then f 0.x0/ 0: D 2 0. From ¤ (2.3.16) Proof We will show that x0 is not a local extreme point of f if f 0.x0/ Lemma 2.3.2, f .x/ x f .x0/ x0 0. Therefore, if f 0.x0/ D f 0.x0/ E.x/; C where limx x0 E.x/ ! D 0, there is a ı > 0 such that ¤ E.x/ j j < f 0.x0/ j j if x j x0 j < ı; and the right side of (2.3.16) must have the same sign as f 0.x0/ for the same is true of the left side, f .x/ x0 (since x interval about x0. < ı. Since f .x0/ must change sign in every neighborhood of x0 does). Therefore, neither (2.3.14) nor (2.3.15) can hold for all x in any x0 x j j If f 0.x0/ 0, we say that x0 is a critical point of f . Theorem 2.3.7 says that every local extreme point of f at which f is differentiable is a critical point of f . The converse is false. For example, 0 is a critical point of f .x/ x3, but not a local extreme point. D D Rolle’s Theorem The use of Theorem 2.3.7 for ﬁnding local extreme points is covered in calculus, so we will not pursue it here. However, we will use Theorem 2.3.7 to prove the following fundamental theorem, which says that if a curve y a and b and has a tangent at .x; f .x// for every x in .a; b/, then there is a point c in .a; b/ x such that the tangent to the curve at .c; f .c// is horizontal (Figure 2.3.4). f .x/ intersects a horizontal line at x D D D 82 Chapter 2 DifferentialCalculusofFunctionsofOneVariable y a c b x Figure 2.3.4 Theorem 2.3.8 (Rolle’s Theorem) Suppose that f is continuous on the closed interval Œa; b and differentiable on the open interval .a; b/; and f .a/ f .b/: Then f 0.c/ 0 for some c in the open interval .a; b/: D D Proof Since f is continuous on Œa; b, f attains a maximum and a minimum value on Œa; b (Theorem 2.2.9). If these two extreme values are the same, then f is constant on 0 for all x in .a; b/. If the extreme values differ, then at least one must .a; b/, so f 0.x/ be attained at some point c in the open interval .a; b/, and f 0.c/ 0, by Theorem 2.3.7. D D Intermediate Values of Derivatives A derivative may exist on an interval Œa; b without being continuous on Œa; b. Nevertheless, an intermediate value theorem similar to Theorem 2.2.10 applies to derivatives. Theorem 2.3.9 (Intermediate Value Theorem for Derivatives) Suppose that f is differentiable on Œa; b; f 0.a/ for some c in .a; b/: f 0.c/ f 0.b/; and is between f 0.a/ and f 0.b/: Then ¤ D Proof Suppose ﬁrst that and deﬁne Then and (2.3.17) implies that f 0.a/ < < f 0.b/ (2.3.17) g.x/ f .x/ x: D g0.x/ f 0.x/ D ; a x b; g0.a/ < 0 and g0.b/ > 0: (2.3.18) (2.3.19) Since g is continuous on Œa; b, g attains a minimum at some point c in Œa; b. Lemma 2.3.2 and (2.3.19) imply that there is a ı > 0 such that g.x/ < g.a/; a < x < a ı; and g.x/ < g.b/; b ı < x < b C Section 2.3 DifferentiableFunctionsofOneVariable 83 (Exercise 2.3.3), and therefore c g0.c/ 0, by Theorem 2.3.7. From (2.3.18), f 0.c/ a and c ¤ ¤ D . D b. Hence, a < c < b, and therefore The proof for the case where f 0.b/ < < f 0.a/ can be obtained by applying this result to f . Mean Value Theorems Theorem 2.3.10 (Generalized Mean Value Theorem) If f and g are continuous on the closed interval Œa; b and differentiable on the open interval .a; b/; then Œg.b/ g.a/f 0.c/ Œf .b/ D f .a/g0.c/ (2.3.20) for some c in .a; b/: Proof The function D is continuous on Œa; b and differentiable on .a; b/, and h.x/ Œg.b/ g.a/f .x/ Œf .b/ f .a/g.x/ h.a/ h.b/ D D g.b/f .a/ f .b/g.a/: Therefore, Rolle’s theorem implies that h0.c/ h0.c/ this implies (2.3.20). Œg.b/ D D g.a/f 0.c/ 0 for some c in .a; b/. Since Œf .b/ f .a/g0.c/; The following special case of Theorem 2.3.10 is important enough to be stated separately. Theorem 2.3.11 (Mean Value Theorem) If f is continuous on the closed interval Œa; b and differentiable on the open interval .a; b/; then for some c in .a; b/: f 0.c/ f .b/ b D f .a/ a Proof Apply Theorem 2.3.10 with g.x/ x. D Theorem 2.3.11 implies that the tangent to the curve y f .x/ at .c; f .c// is parallel to the line connecting the points .a; f .a// and .b; f .b// on the curve (Figure 2.3.5, page 84). D Consequences of the Mean Value Theorem If f is differentiable on .a; b/ and x1, x2 .a; b/ then f is continuous on the closed 2 interval with endpoints x1 and x2 and differentiable on its interior. Hence, the mean value theorem implies that D for some c between x1 and x2. (This is true whether x1 < x2 or x2 < x1.) The next three theorems follow from this. f .x2/ f .x1/ f 0.c/.x2 x1/ 84 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Theorem 2.3.12 If f 0.x/ D 0 for all x in .a; b/; then f is constant on .a; b/: Theorem 2.3.13 If f 0 exists and does not change sign on .a; b/; then f is monotonic on .a; b/ increasing; nondecreasing; decreasing; or nonincreasing as W f 0.x/ > 0; f 0.x/ 0; f 0.x/ < 0; or f 0.x/ 0; respectively; for all x in .a; b/: Theorem 2.3.14 If then f 0.x/ j j M; a < x < b; f .x/ j f .x0/ j M x j ; x0j x; x0 2 .a; b/: (2.3.21) A function that satisﬁes an inequality like (2.3.21) for all x and x0 in an interval is said to satisfy a Lipschitz condition on the interval. y f (b) f (c) f (a) y = f (x) a c b x Figure 2.3.5 2.3 Exercises 1. Prove that a function f is differentiable at x0 if and only if f .x/ f .x0/ x m.x x0 x0/ 0 D lim x0 x ! for some constant m. In this case, f 0.x0/ m. D Section 2.3 DifferentiableFunctionsofOneVariable 85 2. Prove: If f is deﬁned on a neighborhood of x0, then f is differentiable at x0 if and only if the discontinuity of h.x/ f .x/ x D f .x0/ x0 at x0 is removable. 3. Use Lemma 2.3.2 to prove that if f 0.x0/ > 0, there is a ı > 0 such that f .x/ < f .x0/ if x0 ı < x < x0 and f .x/ > f .x0/ if x0 < x < x0 ı: C 4. Suppose that p is continuous on .a; c and differentiable on .a; c/, while q is continuous on Œc; b/ and differentiable on .c; b/. Let f .x/ (a) Show that p.x/; a < x c; D ( q.x/; c < x < b: f 0.x/ p0.x/; a < x < c; D ( q0.x/; c < x < b: (b) Under what additional conditions on p and q does f 0.c/ exist? Prove that Find all derivatives of f .x/ your stated conditions are necessary and sufﬁcient. xn Suppose that f 0.0/ exists and f .x exists for all x. j y/ D C D x 1 j , where n is a positive integer. f .x/f .y/ for all x and y. Prove that f 0 Suppose that c0.0/ a and s0.0/ D D b where a2 b2 0, and c.x s.x y/ y/ C C D D c.x/c.y/ s.x/c.y/ C ¤ C s.x/s.y/ c.x/s.y/ for all x and y. (a) Show that c and s are differentiable on . of c, s, a, and b. ; 1 /, and ﬁnd c0 and s0 in terms 1 (b) (For those who have studied differential equations.) Find c and s explicitly. (a) Suppose that f and g are differentiable at x0, f .x0/ 0. Without using L’Hospital’s rule, show that g.x0/ D D 0, and g0.x0/ ¤ 5. 6. 7. 8. f .x/ g.x/ D f 0.x0/ g0.x0/ : lim x0 x ! (b) State the corresponding results for one-sided limits. Prove Theorem 2.3.4(a). 9. 86 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 10. 11. 12. Prove Theorem 2.3.4(b). Prove Theorem 2.3.4(d). Prove by induction: If n and 1 and f .n/.x0/ and g.n/.x0/ exist, then so does .fg/.n/.x0/, .fg/.n/.x0/ D n 0 m X D n m! m/.x0/: f .m/.x0/g.n HINT: See Exercise 1.2.19: This is Leibniz’s rule for differentiating a product. 13. What is wrong with the “proof” of the chain rule suggested after Example 2.3.3? Correct it. 14. 15. 16. 17. 18. ¤ D 1=f 0.x0/. Suppose that
f is continuous and increasing on Œa; b. Let f be differentiable at a 0. If g is the inverse of f Theorem 2.2.15), show point x0 in .a; b/, with f 0.x0/ that g0.f .x0// (a) Show that f 0 C (b) Example 2.3.4 shows that f 0 C / exists but f 0 C (c) Complete the following statement so it becomes a theorem, and prove the /.” .a/ may exist even if f 0.a .a/ does not. / if both quantities exist. example where f 0.a / does not. Give an / exists and f is theorem: “If f 0.a f 0.a f 0.a .a/ .a/ C D C C at a, then f 0 C D C / and f .b / exist (ﬁnite) if f 0 is bounded on .a; b/. HINT: See C Show that f .a C Exercise 2.1.38: f .a//=.b Suppose that f is continuous on Œa; b, f 0 C .f .b/ Suppose that f is continuous on Œa; b, f 0 C a/. Show that f .c/ .a/ exists, and is between f 0 C f .a/ .a/ and a/ for some c in .a; b/. .c .b/, and D .a/ < < f 0 : a/ .f .b/ Show that either f .c/ in .a; b/. f .a/ D f .a//=.b .c ¤ a/ or f .c/ f .b/ .c D b/ for some c 19. Let f .x/ sin x x ; D x 0: (a) Deﬁne f .0/ so that f is continuous at x (b) Show that if x is a local extreme point of f , then D ¤ 0. HINT: Use Exercise 2.3.8: f .x/ j D j .1 C x2/ 1=2: HINT: Express sin x and cos x in terms of f .x/ and f 0.x/; and add their squares to obtain a useful identity: (c) Show that f .x/ j j 1 for all x. For what value of x is equality attained? Section 2.3 DifferentiableFunctionsofOneVariable 87 20. Let n be a positive integer and f .x/ sin nx n sin x ; D k (k x ¤ D integer): (a) Deﬁne f .k / so that f is continuous at k . HINT: Use Exercise 2.3.8: (b) Show that if x is a local extreme point of f , then f .x/ j D j .n2 1 C 1/ sin2 x 1=2 : HINT: Express sin nx and cos nx in terms of f .x/ and f 0.x/; and add their squares to obtain a useful identity: (c) Show that f .x/ 1 for all x. For what values of x is equality attained? j j 21. We say that f has at least n zeros, counting multiplicities, on an interval I if there are distinct points x1, x2, . . . , xp in I such that f .j /.xi / 0; 0 j ni 1; 1 i p; D np and n1 counting multiplicities, on an interval I , then f 0 has at least n multiplicities, on I . n. Prove: If f is differentiable and has at least n zeros, 1 zeros, counting C C D 22. Give an example of a function f such that f 0 exists on an interval .a; b/ and has a jump discontinuity at a point x0 in .a; b/, or show that there is no such function. 23. Let x1, x2, . . . , xn and y1, y2, . . . , yn be in .a; b/ and yi < xi , 1 that if f is differentiable on .a; b/, then i n. Show n 1 i X D Œf .xi / f .yi / f 0.c/ D n 1 i X D .xi yi / for some c in .a; b/. 24. Prove or give a counterexample: If f is differentiable on a neighborhood of x0, then f satisﬁes a Lipschitz condition on some neighborhood of x0. 25. Let f 00.x/ C p.x/f .x/ D 0 and g00.x/ p.x/g.x/ C D 0; a < x < b: (a) Show that W (b) Prove: If W g.c/ D D f 0g 0 and f .x1/ fg0 is constant on .a; b/. f .x2/ D 0 for some c in .x1; x2/. HINT: Consider f =g: ¤ D 0 where a < x1 < x2 < b, then 88 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 26. Suppose that we extend the deﬁnition of differentiability by saying that f is differentiable at x0 if lim x0 x ! exists in the extended reals. Show that if f 0.x0/ D f .x/ x f .x0/ x0 f .x/ D ( px; 0; x p x; x < 0; 27. . D 1 then f 0.0/ Prove or give a counterexample: If f is differentiable at x0 in the extended sense of Exercise 2.3.26, then f is continuous at x0. 28. Assume that f is differentiable on . ; 1 f .x/g.x/, where g is differentiable on . 1 / and x0 is a critical point of f . (a) Let h.x/ D ; 1 / and 1 Show that the tangent line to the curve y line to the curve y D g.x/ at .x0; g.x0/ intersect on the x-axis. ¤ h.x/ at .x0; h.x0// and the tangent f .x0/g0.x0/ 0: D (b) Suppose that f .x0/ Show that the tangent line to the curve y x-axis at x x1. D D 0. Let h.x/ ¤ D f .x/.x x1/, where x1 is arbitrary. h.x/ at .x0; h.x0// intersects the (c) Suppose that f .x0/ 0. Let h.x/ that the tangent line to the curve y at the midpoint of the interval with endpoints x0 and x1. x1/2, where x1 x0. Show f .x/.x h.x/ at .x0; h.x0// intersects the x-axis D D ¤ ¤ .ax2 (d) Let h.x/ D b . Show that the tangent line to the curve y 2a x0 intersects the x-axis at x x1/, where a D c/.x bx C C ¤ x1. 0 and b2 4ac ¤ 0. Let h.x/ at .x0; h.x0// D (e) Let h be a cubic polynomial with zeros ˛, ˇ, and , where ˛ and ˇ are distinct D ˛ ˇ and is real. Let x0 y D h.x/ at .x0; h.x0// intersects the axis at x C 2 D . D . Show that the tangent line to the curve 2.4 L’HOSPITAL’S RULE The method of Theorem 2.1.4 for ﬁnding limits of the sum, difference, product, and quotient of functions breaks down in connection with indeterminate forms. The generalized mean value theorem (Theorem 2.3.10) leads to a method for evaluating limits of indeterminate forms. Theorem 2.4.1 (L’Hospital’s Rule) Suppose that f and g are differentiable and g0 has no zeros on .a; b/: Let lim b ! x f .x/ D x lim b ! g.x/ 0 D (2.4.1) Section 2.4 L’Hospital’sRule 89 or x and suppose that Then lim b ! f .x/ D ˙1 and x lim b ! g.x/ ; D ˙1 f 0.x/ g0.x/ D lim b ! x L .ﬁnite or /: ˙1 f .x/ g.x/ D L: lim b ! x (2.4.2) (2.4.3) (2.4.4) Proof We prove the theorem for ﬁnite L and leave the case where L (Exercise 2.4.1). to you D ˙1 Suppose that > 0. From (2.4.3), there is an x0 in .a; b/ such that ˇ ˇ ˇ Theorem 2.3.10 implies that if x and t are in Œx0; b/, then there is a c between them, and ˇ therefore in .x0; b/, such that < if x0 < c < b: (2.4.5) f 0.c/ g0.c/ L ˇ ˇ ˇ ˇ Since g0 has no zeros in .a; b/, Theorem 2.3.11 implies that D Œg.x/ g.t/f 0.c/ Œf .x/ f .t/g0.c/: (2.4.6) g.x/ g.t/ ¤ 0 if x; t .a; b/: 2 This means that g cannot have more than one zero in .a; b/. Therefore, we can choose x0 so that, in addition to (2.4.5), g has no zeros in Œx0; b/. Then (2.4.6) can be rewritten as f .x/ g.x/ f .t/ g.t/ D f 0.c/ g0.c/ ; so (2.4.5) implies that f .x/ g.x/ f .t/ g.t < if x; t Œx0; b/: 2 (2.4.7) If (2.4.1) holds, let x be ﬁxed in Œx0; b/, and consider the function From (2.4.1), so t G.t/ f .x/ g.x/ D f .t/ g.t/ L: lim b ! f .t/ D t lim b ! g.t/ 0; D G.t/ lim b ! t f .x/ g.x/ L: D (2.4.8) 90 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Since because of (2.4.7), (2.4.8) implies that G.t/ j j < if x0 < t < b; f .x/ g.x This holds for all x in .x0; b/, which implies (2.4.4). The proof under assumption (2.4.2) is more complicated. Again choose x0 so that (2.4.5) holds and g has no zeros in Œx0; b/. Letting t x0 in (2.4.7), we see that D f .x/ g.x/ f .x0/ g.x0/ < , we can choose x1 > x0 so that f .x/ if x0 x < b2.4.9) 0 and f .x/ f .x0/ ¤ ¤ Since limx if x1 < x < b. Then the function D ˙1 f .x/ ! b is deﬁned and nonzero if x1 < x < b, and u.x/ D g.x0/=g.x/ f .x0/=f .x/ 1 1 lim b ! x u.x/ 1; D (2.4.10) because of (2.4.2). Since f .x/ g.x/ (2.4.9) implies that f .x0/ g.x0/ D f .x/ g.x/ 1 1 f .x0/=f .x/ g.x0/=g.x/ D f .x/ g.x/u.x/ ; f .x/ g.x/u.x/ < if x1 < x < b; u.x/ j j if x1 < x < b: (2.4.11) Lu.x/ L j j u.x/ j C j u.x/ L j j 1 : j (2.4.12) L ˇ ˇ ˇ ˇ < ˇ ˇ which can be rewritten as ˇ ˇ f .x/ g.x/ From this and the triangle inequality, ˇ ˇ ˇ ˇ f .x/ g.x/ Lu.x/ ˇ ˇ ˇ ˇ C j Lu.x/ ˇ ˇ ˇ ˇ and therefore f .x/ g.x.x/ ˇ ˇ ˇ ˇ Because of (2.4.10), there is a point x2 in .x1; b/ such that 1 < This, (2.4.11), and (2.4.12) imply that j j u.x/ j j < 1 C if x2 < x < b: f .x/ g.x1 / L j C j C if x2 < x < b; which proves (2.4.4) under assumption (2.4.2). Section 2.4 L’Hospital’sRule 91 Theorem 2.4.1 and the proof given here remain valid if b replaced by “x that similar theorems are valid for limits from the right, limits at sided) limits. We will take these as given. ” is ” throughout. Only minor changes in the proof are required to show , and ordinary (two- and “x D 1 ! 1 1 ! b The Indeterminate Forms 0=0 and We say that f =g is of the form 0=0 as x b ! or of the form = 1 1 as x ! lim b f .x/ D x ! b if lim b x ! and x lim b ! f .x/ D ˙1 1 = 1 if g.x/ 0; D x The corresponding deﬁnitions for x these forms as x and as x b ! ! lim b ! ! b C g.x/ : D ˙1 and x b C , then we say that it is of that form as x ! ˙1 are similar. If f =g is of one of b. ! Example 2.4.1 The ratio sin x=x is of the form 0=0 as x yields ! 0, and L’Hospital’s rule lim 0 x ! Example 2.4.2 The ratio e rule yields sin x x D lim 0 x ! cos x 1 D 1: x=x is of the form , and L’Hospital’s ! 1 1 as lim !1 Example 2.4.3 Using L’Hospital’s rule may lead to another indeterminate form; thus, lim !1 x x if the limit on the right exists in the extended reals. Applying L’Hospital’s rule again yields ex x2 D ex 2x lim x !1 lim x !1 ex 2x D lim x !1 ex : 2 D 1 lim x !1 ex : x2 D 1 lim x !1 Therefore, More generally, for any real number ˛ (Exercise 2.4.33). lim x !1 ex x˛ D 1 92 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Example 2.4.4 Sometimes it pays to combine L’Hospital’s rule with other manipulations. For example, 4 4 cos x x4 lim 0 x ! 2 sin2 x D D D D D lim 0 x ! lim 0 x ! 4 sin x 4 sin x cos x 4x3 sin x x sin x x lim 0 x ! lim 0 x ! 2 1 cos x x2 sin x 2x lim 0 x ! 1 2 1 2 lim 0 x ! .1/2 D sin x x 1 2 (Example 2.4.1): As another example, L’Hospital’s rule yields x2 e log.1 x x/ C lim 0 x ! D lim 0 x ! x2 2xe log.1 C x/ 1 C x2 e .1 1 x/ C 1: D However, it is better to remove the “determinate” part of the ratio before using L’Hospital’s rule: x2 e x/ log.1 x C lim 0 x ! D D D x/ log.1 C x lim 0 x ! x/ log.1 C x x2 e lim 0 x ! .1/ lim 0 x ! 1=.1 C 1 lim 0 x ! x/ 1: D In using L’Hospital’s rule we usually write, for example, f .x/ g.x/ D f 0.x/ g0.x/ lim b x ! lim b x ! (2.4.13) and then try to ﬁnd the limit on the right. This is convenient, but technically incorrect, since (2.4.13) is true only if the limit on the right exists in the extended reals. It may happen that the limit on the left exists but the one on the right does not. In this case, (2.4.13) is incorrect. Example 2.4.5 If f .x/ x D x2 sin 1 x and g.x/ sin x; D then f 0.x/ 1 D 2x sin 1 x C cos 1 x and g0.x/ cos x: D Section 2.4 L’Hospital’sRule 93 Therefore, limx 0 f 0.x/=g0.x/ does not exis
t. However, ! f .x/ g.x/ D lim 0 x ! lim 0 x ! 1 x sin.1=x/ .sin x/=x D 1 1 D 1: The Indeterminate Form 0 1 We say that a product fg is of the form 0 0 and the other approaches L’Hospital’s rule after writing ˙1 as x 1 b ! b if one of the factors approaches In this case, it may be useful to apply ! as x . f .x/g.x/ f .x/ 1=g.x/ D or f .x/g.x/ g.x/ 1=f .x/ ; D since one of these ratios is of the form 0=0 and the other is of the form Similar statements apply to limits as x , x b C ! ! b, and x as 1 Example 2.4.6 The product x log x is of the form 0 an form yields = 1 1 as x 0 C ! 1 . Converting it to x log x lim 0 C ! x D x lim 0 C ! log x 1=x D x D lim 0 C ! lim 0 C ! x 1=x 1=x2 0: x D Converting to a 0=0 form leads to a more complicated problem: x log x lim 0 C ! x D x lim 0 C ! x 1= log x D x lim 0 C ! 1 1=x.log x/2 D x lim 0 C ! x.log x/2 ‹ D Example 2.4.7 The product x log.1 it to a 0=0 form yields C 1=x/ is of the form 0 as x 1 . Converting ! 1 x log.1 lim x !1 1=x/ C lim x !1 D log.1 1=x/ C 1=x Œ1=.1 C lim x !1 D 1=x/ . 1=x2 1=x2/ lim x !1 1 D 1 1=x D C 1: 94 Chapter 2 DifferentialCalculusofFunctionsofOneVariable In this case, converting to an = 1 1 form complicates the problem: lim x !1 x log.1 1=x/ C lim x !1 lim x !1 D D lim x !1 D x 1= log.1 C 1=x/ 1 1 1=x/2 C 1/Œlog.1 C 1 C 1=x/2 1=x2 1=x ‹ D Œlog.1 x.x C The Indeterminate Form 1 1 1 1 f .x/ as x b ! g.x/ if : D ˙1 A difference f g is of the form lim b ! In this case, it may be possible to manipulate f indeterminate, or is of the form 0=0 or as x lim b ! b, or 1 ! C ! as x g into an expression that is no longer . Similar remarks apply to limits b ! Example 2.4.8 The difference sin x x2 1 x is of the form 1 1 as x ! 0, but it can be rewritten as the 0=0 form sin x x2 x : Hence, sin x x2 1 x lim 0 x ! D D lim 0 x ! lim 0 x ! x sin x x2 sin x 2 D 1 cos x 2x lim 0 x ! D 0: Example 2.4.9 The difference x2 x is of the form as x 1 1 . Rewriting it as ! 1 which is no longer indeterminate as x x2 1 1 x ; , we ﬁnd that ! 1 .x2 lim x !1 x/ lim x !1 D 1 x x2 1 x2 1 lim x !1 1 x lim x !1 /.1/ . 1 D D D 1 Section 2.4 L’Hospital’sRule 95 The Indeterminate Forms 00, 11, and 0 1 The function f g is deﬁned by f .x/g.x/ D eg.x/ log f .x/ D exp.g.x/ log f .x// for all x such that f .x/ > 0. Therefore, if f and g are deﬁned and f .x/ > 0 on an interval .a; b/, Exercise 2.2.22 implies that lim b x Œf .x/g.x/ exp D lim b x g.x/ log f .x/ exists in the extended reals. (If this limit is ! ! g.x/ log f .x/ (2.4.14) 0 and e1 D 1 .) The product g log f can be of the form 0 ˙1 then (2.4.14) is 1 b ! if limx valid if we deﬁne e1 D in three ways as x : b (a) If limx (b) If limx (c) If limx ! b b ! b ! ! g.x/ g.x/ g.x/ D D 0 and limx f .x/ 0. b and limx ! D f .x/ 1. b ! f .x/ D . D 1 b ! D ˙1 0 and limx In these three cases, we say that f g is of the form 00, 11, and b . Similar deﬁnitions apply to limits as x Example 2.4.10 The function xx is of the form 00 as x ! ! , x C b b, and x 1 . ! ˙1 0 C ! . Since 0, respectively, as x ! and limx 0 ! C xx D ex log x x log x D 0 (Example 2.4.6), lim 0 C ! x xx e0 1: D D Example 2.4.11 The function x1=.x 1/ is of the form 11 as x 1. Since ! and it follows that x1=.x 1/ exp D log x 1 x lim 1 x ! log x x 1 D 1=x 1 D 1; lim 1 x ! x1=.x 1/ e1 e: D D lim 1 x ! Example 2.4.12 The function x1=x is of the form 0 as x 1 . Since ! 1 and x1=x exp D log x x lim x !1 log x x D lim x !1 1=x 1 D 0; 96 Chapter 2 DifferentialCalculusofFunctionsofOneVariable it follows that lim x !1 x1=x e0 1: D D 2.4 Exercises 1. Prove Theorem 2.4.1 for the case where limx b ! f 0.x/=g0.x/ . D ˙1 In Exercises 2.4.2–2.4.40, ﬁnd the indicated limits. 2. 5. 8. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. tan sin 1 x 1 x lim 0 x ! sin nx sin x x sin.1=x/ lim x ! lim x !1 3. 6. 9. lim x ! sin x log. j tan x / j lim x !1 .px 1 C px/ lim 0 x ! .cot x csc x/ tan x lim x ! j sin x j lim x ! 0 j sin x x j xsin.1=x/ lim x !1 x˛ log x lim 0 C ! x px2 1 lim x !1 1 1 x x C .log x/ˇ x lim x !1 lim x !1 .x˛ log x/ lim 0 x ! 1 log.1 cos x x2/ C x/ log.1 C x lim 0 x ! 4. 7. lim x !1 px.e 1=x 1/ 10. lim 0 C ! x 1 C ex cos x 1 ex sin e x2 lim x !1 tan x log x lim 0 C ! x 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. log.tan x/ 1 x 1 1 x cos x 1 x C 1 ex 1 x lim 0 C ! lim 0 x ! lim 0 x ! sin x lim ! =2 j x tan x j x/1=x .1 lim 0 x ! C lim 0 x ! lim e x ! 1 log.log x/ e/ sin.x x cos x 2 x x lim 1 C ! x x 1 1 C px2 1 .cosh x sinh x/ sin.ex/ ex2 lim x !1 lim !1 x 31. 33. 35. 37. 38. 39. 40. 41. Section 2.4 L’Hospital’sRule 97 x.x lim x !1 C 1/ Œlog.1 C 1=x/2 32. sin x x x5 C lim 0 x ! x3=6 34. 36. etan x cos x lim 3=2 x ! lim x !1 xx x log x ex x˛ lim x !1 lim 1 ! C lim =2 x x ! .log x/˛ log.log x/ .sin x/tan x n xr r Š ex 0 r X D xn .n integer 1/ lim 0 x ! lim 0 x ! sin x 1=x2 e n D 1/r x2r . .2r 0 r C X D x2n 1 C 1 C lim D x 0 ! (a) Prove: If f is continuous at x0 and limx xn D integer) (n 0 and f 0 is continuous at x0. 1/Š .n D integer 0/ x0 f 0.x/ exists, then f 0.x0/ exists ! (b) Give an example to show that it is necessary to assume in (a) that f is con- tinuous at x0. 42. The iterated logarithms are deﬁned by L0.x/ x and D x > an; 1; n Ln.x/ log.Ln 1.x//; D ean1; n where a1 D (a) Ln.x/ (b) Ln (c) lim an ! lim x !1 (d) D 1.an D 1.log x/; 0 and an Ln / D 1.x//˛ Ln.x/ C .Ln.x//˛ =Ln C .Ln 1.x/ x 0 and Ln.an D D 1. Show that x > an; / n 1. . C D 1 0 if ˛ > 0 and n 1. 43. Let f be positive and differentiable on .0; 0 if ˛ is arbitrary and n 1. /, and suppose that 1 L; where 0 < L : 1 f 0.x/ f .x/ D lim x !1 Deﬁne f0.x/ x and D D Use L’Hospital’s rule to show that fn.x/ f .fn 1.x// ; 1: n lim x !1 .fn.x//˛ fn 1.x/ D 1 if ˛ > 0 and n 1: 98 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 44. Let f be differentiable on some deleted neighborhood N of x0, and suppose that f and f 0 have no zeros in N . Find f .x/ if lim (a) f .x/ j x0 ! 1/ if lim f .x/ j x x0 ! 1=f .x/ if limx x0 f .x/ D f .x/ f .x/ 1=.f .x/ (b) (c) x0 j x0 j 0; x f .x/ j x0 j lim x ! lim x ! lim x ! ! 1; D . D 1 45. Suppose that f and g are differentiable and g0 has no zeros on .a; b/. Suppose also that limx L and either f 0.x/=g0.x/ b ! D lim b ! x f .x/ D x lim b ! g.x/ 0 D or Find limx .1 b ! C 46. We distinguish between / and /, 1 1 and . D 1 . 0 1 1 1 x x and g.x/ f .x/ D 1 lim b ! : D ˙1 lim b ! f .x//1=g.x/. / and between . . 1 1 D 1 D 1 /. Why don’t we distinguish between 0 . D 1 , and 11 and 11 / and . = 1 and 1 1 1 C 1 and 1 2.5 TAYLOR’S THEOREM A polynomial is a function of the form p.x/ a0 D C a1.x x0/ C C an.x x0/n; (2.5.1) an The polynomial (2.5.1) is said to be written in powers of x 0. If we wish to leave open the possibility that an where a0, . . . , an and x0 are constants. Since it is easy to calculate the values of a polynomial, considerable effort has been devoted to using them to approximate more complicated functions. Taylor’s theorem is one of the oldest and most important results on this question. x0, and is of degree n if 0, we say that p is of degree 0. If 0, so that p vanishes identically, then p has no degree according to our deﬁnition, a0 which requires at least one coefﬁcient to be nonzero. For convenience we say that the identically zero polynomial p has degree . (Any negative number would do as well as . The point is that with this convention, the statement that p is a polynomial of degree ¤ n. In particular, a constant polynomial p.x/ a0 is of degree zero if a0 1 D D D ¤ n includes the possibility that p is identically zero.) 1 Taylor Polynomials We saw in Lemma 2.3.2 that if f is differentiable at x0, then f .x/ D f .x0/ C f 0.x0/.x x0/ C E.x/.x x0/; where Section 2.5 Taylor’sTheorem 99 E.x/ 0: D lim x0 x ! To generalize this result, we ﬁrst restate it: the polynomial which is of degree T1.x/ D 1 and satisﬁes f .x0/ C f 0.x0/.x x0/; D approximates f so well near x0 that T1.x0/ f .x0/; T 01.x0/ f 0.x0/; D lim x0 x ! f .x/ x T1.x/ x0 0: D (2.5.2) Now suppose that f has n derivatives at x0 and Tn is the polynomial of degree n such that T .r / n .x0/ How well does Tn approximate f near x0? f .r /.x0/; D 0 r n: (2.5.3) To answer this question, we must ﬁrst ﬁnd Tn. Since Tn is a polynomial of degree it can be written as n, Tn.x/ a0 C D a1.x x0/ C C an.x x0/n; (2.5.4) where a0, . . . , an are constants. Differentiating (2.5.4) yields so (2.5.3) determines ar uniquely as T .r / n .x0/ r Šar ; 0 r n; D f .r /.x0/ r Š ; ar D 0 r n: Therefore, Tn.x/ D D .x x0/ C C f .n/.x0/ nŠ .x x0/n f 0.x0/ 1Š C f .r /.x0/ r Š .x f .x0/ n 0 r X D x0/r : We call Tn the nth Taylor polynomial of f about x0. The following theorem describes how Tn approximates f near x0. Theorem 2.5.1 If f .n/.x0/ exists for some integer n polynomial of f about x0; then 1 and Tn is the nth Taylor lim x0 x ! f .x/ .x Tn.x/ x0/n D 0: (2.5.5) 100 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Proof The proof is by induction. Let Pn be the assertion of the theorem. From (2.5.2) we know that (2.5.5) is true if n 1; that is, P1 is true. Now suppose that Pn is true for some integer n 1/ exists. Since the ratio 1, and f .n D C is indeterminate of the form 0=0 as x f .x/ .x Tn C x0/n 1.x/ 1 C x0, L’Hospital’s rule implies that 1.x/ 1 lim x0 x ! 1 f 0.x/ .x T 0n C x0/n ! 1.x/ 1 D n lim x0 x ! f .x/ .x Tn C x0/n C if the limit on the right exists. But f 0 has an nth derivative at x0, and C (2.5.6) T 0n 1.x/ C D n f .r 1/.x0/ C r Š .x x0/r 0 r X D is the nth Taylor polynomial of f 0 about x0. Therefore, the induction assumption, applied to f 0, implies that This and (2.5.6) imply that lim x0 x ! lim x0 x ! which completes the induction. It can be shown (Exercise 2.5.8) that if f 0.x/ .x 1.x/ T 0n C x0/n 0: D f .x/ .x Tn C x0/n 1.x/ 1 D C 0; pn is a polynomial of degree D then a0 a1.x C n such that x0/ C C an.x x0/n lim x0 x ! f .x/ .x pn.x/ x0/n D 0; f .r /.x0/ r Š I ar D that is, pn x0 in the manner indicated in (2.5.5). D Tn. Thus, Tn is the only polynomial of degree n that approximates f near Theorem 2.5.1 can be restated as a generalization of Lemma
2.3.2. Lemma 2.5.2 If f .n/.x0/ exists; then f .x/ n D 0 r X D f .r /.x0/ r Š .x x0/r C En.x/.x x0/n; (2.5.7) where lim x0 x ! En.x/ En.x0/ 0: D D Proof Deﬁne En.x/ f .x/ .x Tn.x/ x0/n D 8 < 0; Section 2.5 Taylor’sTheorem 101 x0 ; g f ; x x Df x0: 2 D Then (2.5.5) implies that limx (2.5.7). x0 En.x/ : ! D En.x0/ D 0, and it is straightforward to verify Example 2.5.1 If f .x/ so the nth Taylor polynomial of f about x0 ex, then f .n/.x/ D D 0 is D ex. Therefore, f .n/.0/ 1 for n 0, D Tn.x/ n D 0 r X D xr r Š D 1 C x 1Š C x2 2Š C C xn nŠ : (2.5.8) Theorem 2.5.1 implies that (See also Exercise 2.4.38.) ex xr r Š n 0 r X D xn lim 0 x ! 0: D Example 2.5.2 If f .x/ log x, then f .1/ 0 and D D f .r /.x/ 1/.r . D 1/ .r 1/Š xr ; r 1; so the nth Taylor polynomial of f about x0 1 is D Tn.x/r r 1/r .x if n 1. (T0 D 0.) Theorem 2.5.1 implies that n log x lim 1 x ! . 1/r 1r .x 1/r 1 r X D .x 1/n 0; n 1: D Example 2.5.3 If f .x/ D .1 C q.1 q.q x/q , then 1 x/q 1/.1 C C x/q 2 f 0.x/ f 00.x/ f .n/.x/ D D ::: D q.q 1/ .q n C 1/.1 C x/q n: 102 Chapter 2 DifferentialCalculusofFunctionsofOneVariable If we deﬁne then q 0! D 1 and q n! D q.q 1/ .q nŠ n 1/ ; C 1; n and the nth Taylor polynomial of f about 0 can be written as f .n/.0/ nŠ D q n! ; Tn.x/ n D 0 r X D xr : q r ! (2.5.9) Theorem 2.5.1 implies that n x/q .1 C lim 0 x ! 0 r X D xn xr q r ! 0; n 0: D q n! If q is a nonnegative integer, then is the binomial coefﬁcient deﬁned in Exer- cise 1.2.19. In this case, we see from (2.5.9) that Tn.x/ .1 C D x/q D f .x/; n q: Applications to Finding Local Extrema Lemma 2.5.2 yields the following theorem. Theorem 2.5.3 Suppose that f has n derivatives at x0 and n is the smallest positive integer such that f .n/.x0/ (a) If n is odd; x0 is not a local extreme point of f: (b) If n is even; x0 is a local maximum of f if f .n/.x0/ < 0; or a local mininum of f if ¤ 0: f .n/.x0/ > 0: Proof Since f .r /.x0/ 0 for 1 D r n 1, (2.5.7) implies that f .x/ f .x0/ D " En.x/ # C x0/n .x (2.5.10) f .n/.x0/ nŠ in some interval containing x0. Since limx ı > 0 such that x0 En.x/ 0 and f .n/.x0/ D 0, there is a ¤ ! f .n/.x0/ nŠ ˇ ˇ ˇ ˇ ˇ En.x/ j j < ˇ ˇ ˇ ˇ ˇ if x j x0 j < ı: This and (2.5.10) imply that Section 2.5 Taylor’sTheorem 103 (2.5.11) f .x/ .x f .x0/ x0/n x0 x has the same sign as f .n/.x0/ if 0 < < ı. If n is odd the denominator of (2.5.11) j changes sign in every neighborhood of x0, and therefore so must the numerator (since the ratio has constant sign for 0 < < ı). Consequently, f .x0/ cannot be a local extreme value of f . This proves (a). If n is even, the denominator of (2.5.11) is positive for x < ı. This proves (b). f .x0/ must have the same sign as f .n/.x0/ for 0 < x0, so f .x/ x0 x0 ¤ x x j j j j j For n D 2,(b) is called the second derivative test for local extreme points. Example 2.5.4 If f .x/ of f . Since D ex3 , then f 0.x/ 3x2ex3 D , and 0 is the only critical point and f 00.x/ .6x D C 9x4/ex3 f 000.x/ .6 C D 54x3 C 27x6/ex3 ; f 00.0/ point of f . Since f is differentiable everywhere, it has no local maxima or minima. 0. Therefore, Theorem 2.5.3 implies that 0 is not a local extreme 0 and f 000.0/ ¤ D Example 2.5.5 If f .x/ 1=2/, k .k are 0 and ˙ C D D f 00.x/ p f 00.0/ sin x2, then f 0.x/ 0; 1; 2; : : : . Since 2x cos x2, so the critical points of f D 2 cos x2 D 4x2 sin x2; 2 and f 00 D .k 1=2/ / ˙ C 1/k 1.4k C . D 2/: C Therefore, Theorem 2.5.3 implies that f attains local minima at 0 and odd integers k, and local maxima at ˙ 1=2/ for even integers k. p .k ˙ C .k C 1=2/ for p Taylor’s theorem p Theorem 2.5.1 implies that the error in approximating f .x/ by Tn.x/ approaches zero x0/n as x approaches x0; however, it gives no estimate of the error in faster than .x approximating f .x/ by Tn.x/ for a ﬁxed x. For instance, it provides no estimate of the error in the approximation e0:1 T2.0:1/ 1 C D 0:1 1Š C .0:1/2 2Š D 1:105 (2.5.12) obtained by setting n of estimating errors of this kind under the additional assumption that f .n neighborhood of x0. 0:1 in (2.5.8). The following theorem provides a way 1/ exists in a 2 and x D D C 104 Chapter 2 DifferentialCalculusofFunctionsofOneVariable Theorem 2.5.4 (Taylor’s Theorem) Suppose that f .n terval I about x0; and let x be in I: Then the remainder C 1/ exists on an open in- Rn.x/ f .x/ D Tn.x/ can be written as Rn.x/ D f .n .n 1/.c/ 1/Š C C x0/n C 1; .x where c depends upon x and is between x and x0: This theorem follows from an extension of the mean value theorem that we will prove below. For now, let us assume that Theorem 2.5.4 is correct, and apply it. Example 2.5.6 If f .x/ implies that D ex, then f 000.x/ D ex, and Theorem 2.5.4 with n 2 D where c is between 0 and x. Hence, from (2.5.12), ex x 1 C C D x2 2Š C ecx3 3Š ; e0:1 D 1:105 ec.0:1/3 6 ; C where 0 < c < 0:1. Since 0 < ec < e0:1, we know from this that 1:105 < e0:1 < 1:105 e0:1.0:1/3 6 : C The second inequality implies that so Therefore, e0:1 1 .0:1/3 6 < 1:105; e0:1 < 1:1052: 1:105 < e0:1 < 1:1052; and the error in (2.5.12) is less than 0:0002. Example 2.5.7 In numerical analysis, forward differences are used to approximate derivatives. If h > 0, the ﬁrst and second forward differences with spacing h are deﬁned by f .x/ f .x h/ C D f .x/ and 2f .x/ D D Higher forward differences are deﬁned inductively (Exercise 2.5.18). h/ D C C C f .x/: f .x f .x h/ C 2h/ f .x/ 2f .x Œf .x/ (2.5.13) We will ﬁnd upper bounds for the magnitudes of the errors in the approximations Section 2.5 Taylor’sTheorem 105 and f 0.x0/ f .x0/ h f 00.x0/ 2f .x0/ h2 : (2.5.14) (2.5.15) If f 00 exists on an open interval containing x0 and x0 estimate the error in (2.5.14) by writing C h, we can use Theorem 2.5.4 to (2.5.16) f .x0 h/ C D f .x0/ C f 0.x0/h h. We can rewrite (2.5.16) as f 00.c/h2 2 ; C f .x0 C h/ h f .x0/ f 0.x0/ f 0.c/h 2 ; D f .x0/ h f 0.x0/ f 00.c/h 2 : D where x0 < c < x0 C which is equivalent to Therefore, ˇ ˇ where M2 is an upper bound for ˇ ˇ If f 000 exists on an open interval containing x0 and x0 on .x0; x0 j f .x0/ h f 00j f 0.x0/ ˇ ˇ ˇ C ˇ h/. M2h 2 ; to estimate the error in (2.5.15) by writing 2h, we can use Theorem 2.5.4 C f .x0 h/ C D f .x0/ C hf 0.x0/ h2 2 C f 00.x0/ h3 6 C f 000.c0/ and f .x0 C where x0 < c0 < x0 2h/ D f .x0/ C 2hf 0.x0/ h and x0 < c1 < x0 C f .x0 2h/ C 2f .x0 h/ C C f .x0/ D which can be rewritten as 2h2f 00.x0/ f 000.c1/; C 2h. These two equations imply that C 4h3 3 C h2f 00.x0/ 4 3 C f 000.c1/ 1 3 h3; f 000.c0/ 2f .x0/ h2 f 00.x0/ 4 3 D f 000.c1/ 1 3 h; f 000.c0/ because of (2.5.13). Therefore, where M3 is an upper bound for 2f .x0/ h2 f 000j j ˇ ˇ ˇ ˇ f 00.x0/ ˇ ˇ ˇ ˇ C on .x0; x0 5M3h 3 ; 2h/. 106 Chapter 2 DifferentialCalculusofFunctionsofOneVariable The Extended Mean Value Theorem We now consider the extended mean value theorem, which implies Theorem 2.5.4 (Exercise 2.5.24). In the following theorem, a and b are the endpoints of an interval, but we do not assume that a < b. Theorem 2.5.5 (Extended Mean Value Theorem) Suppose that f is continuous on a ﬁnite closed interval I with endpoints a and b .that is, either I .a; b/ or 1/ exists on the open interval I 0; and; if n > 0; that f 0, . . . , f .n/ exist I C and are continuous at a: Then .b; a//; f .n D D f .b/ n 0 r X D f .r /.a/ r Š .b a/r D for some c in I 0: f .n .n 1/.c/ 1/Š C C a/n C 1 .b (2.5.17) Proof The proof is by induction. The mean value theorem (Theorem 2.3.11) implies the conclusion for n 1, and assume that the assertion of the theorem is true with n replaced by n 1. The left side of (2.5.17) can be written as 0. Now suppose that n D f .r /.a/ r Š f .b/ n 0 r X D a/r .b K D .b .n 1 a/n C 1/Š C (2.5.18) for some number K. We must prove that K consider the auxiliary function D f .n C 1/.c/ for some c in I 0. To this end, h.x/ D f .x/ n 0 r X D f .r /.a/ r Š a/r .x K which satisﬁes .x .n 1 a/n C 1/Š ; C (the latter because of (2.5.18)) and is continuous on the closed interval I and differentiable on I 0, with h.a/ 0; h.b/ 0; D D h0.x/ f 0.x/ D f .r 1/.a/ C r Š a/r .x .x K a/n nŠ : (2.5.19) Therefore, Rolle’s theorem (Theorem 2.3.8) implies that h0.b1/ thus, from (2.5.19), 0 for some b1 in I 0; D f 0.b1/ n 1 f .r 1/.a/ C r Š a/r .b1 K a/n .b1 nŠ 0: D 0 r X D If we temporarily write f 0 D g, this becomes g.b1/ n 1 0 r X D g.r /.a/ r a/r .b1 K .b1 a/n nŠ 0: D (2.5.20) n 1 0 r X D Section 2.5 Taylor’sTheorem 107 2 I 0, the hypotheses on f imply that g is continuous on the closed interval J Since b1 with endpoints a and b1, g.n/ exists on J 0, and, if n 1/ exist and are 1, g0, . . . , g.n continuous at a (also at b1, but this is not important). The induction hypothesis, applied to g on the interval J , implies that g.b1/ n 1 0 r X D g.r /.a/ r Š a/r .b1 D g.n/.c/ nŠ a/n .b1 for some c in J 0. Comparing this with (2.5.20) and recalling that g f 0 yields D D Since c is in I 0, this completes the induction. D K g.n/.c/ f .n C 1/.c/: 2.5 Exercises 1. Let f .x/ D 1=x2 e 0; 0; 0: ; x x ¤ D ; 1 Show that f has derivatives of all orders on . of f about 0 is identically zero. HINT: See Exer ci se 2.4.40: Suppose that f .n x0. Show that the function 1 C 2. 1/.x0/ exists, and let Tn be the nth Taylor polynomial of f about / and every Taylor polynomial En.x/ f .x/ .x Tn.x/ x0/n D 8 < 0; x0 ; g f ; x x Df x0; 2 D is differentiable at x0, and ﬁnd E 0n.x0/. (a) Prove: If f is continuous at x0 and there are constants a0 and a1 such that : 3. f .x/ a0 x lim x0 x ! x0/ a1.x x0 0; D then a0 f .x0/, f 0 is differentiable at x0, and f 0.x0/ a1. D D (b) Give a counterexample to the following statement: If f and f 0 are continuous at x0 and there are constants a0, a1, and a2 such that f .x/ a0 lim x0 x ! then f 00.x0/ exists. 4. (a) Prove: if f 00.x0/ exists, then a1.x .x x0/ x0/2 a2.x x0/2 0; D f .x0 h/ C 2f .x0/ h2 C f .x0 h/ f 00.x0/: D lim 0 h ! 108 Chapter 2 DifferentialCalculusofFunctionsofOneVariable (b) Prove or give a counterexample: If the limit in (a) exists, then so does f 00.x0/, and they are equal. 5. A function f has a simple zero (or a zero of multiplicity
1) at x0 if f is differentiable in a neighborhood of x0 and f .x0/ ¤ (a) Prove that f has a simple zero at x0 if and only if 0, while f 0.x0/ D 0. f .x/ D g.x/.x x0/; where g is continuous at x0 and differentiable on a deleted neighborhood of x0, and g.x0/ 0. ¤ (b) Give an example showing that g in(a) need not be differentiable at x0. 6. A function f has a double zero (or a zero of multiplicity 2) at x0 if f is twice dif0. 0, while f 00.x0/ f 0.x0/ ferentiable on a neighborhood of x0 and f .x0/ (a) Prove that f has a double zero at x0 if and only if D ¤ D f .x/ D g.x/.x x0/2; where g is continuous at x0 and twice differentiable on a deleted neighborhood of x0, g.x0/ 0, and ¤ .x lim x0 x ! x0/g0.x/ 0: D (b) Give an example showing that g in(a) need not be differentiable at x0. 7. Let n be a positive integer. A function f has a zero of multiplicity n at x0 if f is n times differentiable on a neighborhood of x0, f .x0/ f .n 1/.x0/ x0 if and only if 0 and f .n/.x0/ D ¤ D D 0. Prove that f has a zero of multiplicity n at D f 0.x0/ f .x/ D g.x/.x x0/n; where g is continuous at x0 and n times differentiable on a deleted neighborhood of x0, g.x0/ 0, and ¤ lim x0 x ! HINT: Use Exercise 2.5.6 and induction: x0/j g.j /.x/ .x 0; 1 j 1: n D 8. (a) Let Q.x/ ˛0 D be a polynomial of degree ˛1.x x0/ n such that C C C ˛n.x x0/n lim x0 x ! Q.x/ x0/n D .x 0: Show that ˛0 ˛1 D D D ˛n 0. D 9. 10. 11. Section 2.5 Taylor’sTheorem 109 (b) Suppose that f is n times differentiable at x0 and p is a polynomial p.x/ D n such that of degree a0 C a1.x x0/ C C an.x x0/n lim x0 x ! f .x/ .x p.x/ x0/n D 0: f .r /.x0/ r Š ar D if 0 r n I Tn, the nth Taylor polynomial of f about x0. Show that that is, p D Show that if f .n/.x0/ and g.n/.x0/ exist and lim x0 x ! f .x/ .x g.x/ x0/n D 0; r n. D g.r /.x0/, 0 then f .r /.x0/ (a) Let Fn, Gn, and Hn be the nth Taylor polynomials about x0 of f , g, and fg. Show that Hn can be obtained by multiplying Fn x0 through the nth. HINT: Use their product h by Gn and retaining only the powers of x Exercise 2.5.8.b/: D (b) Use the method suggested by (a) to compute h.r /.x0/, r (i) h.x/ (ii) h.x/ (iii) h.x/ (iv) h.x/ D D D D D x0 ex sin x; 0 .cos x=2/.log x/; x2 cos x; x/ .1 D =2 x0 (a) It can be shown that if g is n times differentiable at x and f is n times diff .g.x// is n times ferentiable at g.x/, then the composite function h.x/ differentiable at x and x0 x; 1e C D D 0 x0 1 D 1; 2; 3; 4. D h.n/.x/ n D 1 r X D f .r /.g.x// r Š rnŠ r1Š r X r1 g0.x/ 1Š g00.x/ 2Š r2 rn g.n/.x/ nŠ ! where r is over all n-tuples .r1; r2; : : : ; rn/ of nonnegative integers such that P and r1 r2 C C C rn r D r1 2r2 nrn n: C C C (This is Faa di Bruno’s formula). However, this formula is quite complicated. Justify the following alternative method for computing the derivatives of a composite function at a point x0: D 110 Chapter 2 DifferentialCalculusofFunctionsofOneVariable g.x0/, and let Gn and Let Fn be the nth Taylor polynomial of f about y0 Hn be the nth Taylor polynomials of g and h about x0. Show that Hn can be obtained by substituting Gn into Fn and retaining only powers of x x0 through the nth. HINT: See Exercise 2.5.8.b/: D (b) Compute the ﬁrst four derivatives of h.x/ method suggested by (a). cos.sin x/ at x0 0, using the D D 12. (a) If g.x0/ ¤ 0 and g.n/.x0/ exists, then the reciprocal h 1=g is also n times differentiable at x0, by Exercise 2.5.11(a), with f .x/ 1=x. Let Gn and Hn be the nth Taylor polynomials of g and h about x0. Use Exercise 2.5.11(a) to 1, then Hn can be obtained by expanding the polynomial prove that if g.x0/ D D D (b) Use the method of (a) to compute the ﬁrst four derivatives of the following 1 ; (c) Use Exercise 2.5.10 to justify the following alternative procedure for obtaining 1: If Hn, again assuming that g.x0/ x0 C D 0 n 1 r X D Gn.x/r Œ1 x0 and retaining only powers through the nth. in powers of x functions at x0. (i) h.x/ (ii) h.x/ (iii) h.x/ (iv) h.x/ D D D D csc x; .1 x C sec x; Œ1 x0 D x2/ C x0 log.1 =2 1; =4 x/ D C x0 0 D D a1.x Gn.x/ 1 C D x0/ C C an.x x0/n (where, of course, ar Hn.x/ D D then g.r /.x0/=r Š/ and x0/ C C bn.x x0/n; b0 C b1.x k b0 D 1; bk D 1 r X D ar bk r ; 1 k n: 13. Determine whether x0 0 is a local maximum, local minimum, or neither. D x2ex3 x2 x3 1 C 1 C x2 sin3 x ex sin x2 x2 cos x C (a) f .x/ (c) f .x/ (e) f .x/ (g) f .x/ D D D D mum point. (b) f .x/ (d) f .x/ (f ) f .x/ (h) f .x/ D D D D x3ex2 1 C 1 C ex2 x3 x2 sin x ex2 cos x 14. Give an example of a function that has zero derivatives of all orders at a local mini- Section 2.5 Taylor’sTheorem 111 15. Find the critical points of f .x/ x3 3 C bx2 2 C D cx d C and identify them as local maxima, local minima, or neither. 16. Find an upper bound for the magnitude of the error in the approximation. (a) sin x x; x j j (b) p1 x C 1 C < x 2 ; (c) cos x p2 20 x j j < 1 1 1/ x .x ; 4 1//3 .x 16 3 ; x j 1 j < 1 64 (d) log x .x 17. Prove: If then Tn.x/ n D 0 r X D xr r Š ; Tn.x/ < Tn 1.x/ < ex < C 1 if 0 < x < Œ.n C 1/Š1=.n 1/. C 1 xn C 1 .n 1/Š C Tn.x/ 18. The forward difference operators with spacing h > 0 are deﬁned by 0f .x/ D f .x/; f .x/ f .x h/ C D n C 1f .x/ Œnf .x/ ; n D f .x/; 1: (a) Prove by induction on n: If k 2, c1, . . . , ck are constants, and n 1, then nŒc1f1.x/ C C c1nf1.x/ D C C cknfk .x/: ckfk .x/ (b) Prove by induction: If n 1, then n nf .x/ D 0 m X D HINT: See Exercise 1.2.19: 1/n m . n m! f .x C mh/: In Exercises 2.5.19–2.5.22, is the forward difference operator with spacing h > 0. 112 Chapter 2 DifferentialCalculusofFunctionsofOneVariable 19. Let m and n be nonnegative integers, and let x0 be any real number. Prove by induction on n that n.x Does this suggest an analogy between “differencing" and differentiation? x0/m D 0 nŠhn if 0 if m n; m n: D 20. Find an upper bound for the magnitude of the error in the approximation f 00.x0/ 2f .x0 h2 h/ ; (a) assuming that f 000 is bounded on .x0 (b) assuming that f .4/ is bounded on .x0 h; x0 h; x0 C C h/; h/. 21. Let f 000 be bounded on an open interval containing x0 and x0 k such that the magnitude of the error in the approximation f 0.x0/ f .x0/ h C k 2f .x0/ h2 2h. Find a constant C 22. 23. is not greater than M h2, where M Prove: If f .n . j 1/ is bounded on an open interval containing x0 and x0 ˚ x0 < c < x0 f 000.c/ j sup D C j ˇ ˇ nh, then C ˇ ˇ ˇ where An is a constant independent of f and ˇ f .n/.x0/ ˇ ˇ ˇ ˇ nf .x0/ hn AnMn 1h; C Mn 1 C sup x0<c<x0 D nh j C f .n C 1/.c/ j : HINT: See Exercises 2.5.18 and 2.5.19: Suppose that f .n mial of degree then 1/ exists on .a; b/, x0, . . . , xn are in .a; b/, and p is the polynoC .a; b/, n such that p.xi / n. Prove: If x f .xi /, 0 i 2 f .x/ p.x/ C D .x x0/.x x1/ .x xn/; where c, which depends on x, is in .a; b/. HINT: Let x be ﬁxed; distinct from x0; x1; . . . , xn; and consider the function D 1/.c/ 1/Š f .n .n C C g.y/ f .y/ p.y/ .n D K C .y 1/Š x0/.y x1/ .y xn/; where K is chosen so that g.x/ 1/.c/ for some c in .a; b/: f .n 24. Deduce Theorem 2.5.4 from Theorem 2.5.5. D C 0: Use Rolle’s theorem to show that K D CHAPTER 3 Integral Calculus of Functions of One Variable IN THIS CHAPTER we discuss the Riemann on a ﬁnite interval Œa; b, and improper integrals in which either the function or the interval of integration is unbounded. SECTION 3.1 begins with the deﬁnition of the Riemann integral and presents the geometrical interpretation of the Riemann integral as the area under a curve. We show that an unbounded function cannot be Riemann integrable. Then we deﬁne upper and lower sums and upper and lower integrals of a bounded function. The section concludes with the deﬁnition of the Riemann–Stieltjes integral. j j f SECTION 3.2 presents necessary and sufﬁcient conditions for the existence of the Riemann integral in terms of upper and lower sums and upper and lower integrals. We show that continuous functions and bounded monotonic functions are Riemann integrable. SECTION 3.3 begins with proofs that the sum and product of Riemann integrable functions is Riemann integrable if f is Riemann integrable. Other topics are integrable, and that covered include the ﬁrst mean value theorem for integrals, antiderivatives, the fundamental theorem of calculus, change of variables, integration by parts, and the second mean value theorem for integrals. SECTION 3.4 presents a comprehensive discussion of improper integrals. Concepts deﬁned and considered include absolute and conditional convergence of an improper integral, Dirichlet’s test, and change of variable in an improper integral. SECTION 3.5 deﬁnes the notion of a set with Lebesgue measure zero, and presents a necessary and sufﬁcient condition for a bounded function f to be Riemann integrable on an interval Œa; b; namely, that the discontinuities of f form a set with Lebesgue masure zero. 3.1 DEFINITION OF THE INTEGRAL The integral that you studied in calculus is the Riemann integral, named after the German mathematician Bernhard Riemann, who provided a rigorous formulation of the integral to 113 114 Chapter 3 IntegralCalculusofFunctionsofOneVariable replace the intuitive notion due to Newton and Leibniz. Since Riemann’s time, other kinds of integrals have been deﬁned and studied; however, they are all generalizations of the Riemann integral, and it is hardly possible to understand them or appreciate the reasons for developing them without a thorough understanding of the Riemann integral. In this section we deal with functions deﬁned on a ﬁnite interval Œa; b. A partition of Œa; b is a set of subintervals where Thus, any set of n denote by C Œx0; x1; Œx1; x2; : : : ; Œxn 1; xn; a D x0 < x1 < xn b: D (3.1.1) (3.1.2) 1 points satisfying (3.1.2) deﬁnes a partition P of Œa; b, which we P D f x0; x1; : : : ; xn : g The points x0, x1, . . . , xn are the partition points of P . The largest of the lengths of the subintervals (3.1.1) is the norm of P , written as ; thus, P k k P k k D .xi max n i 1 xi 1/: If P and P 0 are partitions of Œa; b, then P 0 is a reﬁnement of P if every par
tition point of P is also a partition point of P 0; that is, if P 0 is obtained by inserting additional points between those of P . If f is deﬁned on Œa; b, then a sum D n 1 j X D f .cj /.xj xj 1/; where xj 1 cj xj ; 1 j n; is a Riemann sum of f over the partition P . (Occasionally we will say D f more simply that is a Riemann sum of f over Œa; b.) Since cj can be chosen arbitrarily in Œxj ; xj 1, there are inﬁnitely many Riemann sums for a given function f over a given partition P . x0; x1; : : : ; xn g Deﬁnition 3.1.1 Let f be deﬁned on Œa; b. We say that f is Riemann integrable on Œa; b if there is a number L with the following property: For every > 0, there is a ı > 0 such that if is any Riemann sum of f over a partition P of Œa; b such that we say that L is the Riemann integral of f over Œa; b, and write P k k < ı. In this casex/ dx L: D Section 3.1 DeﬁnitionoftheIntegral 115 We leave it to you (Exercise 3.1.1) to show that b a f .x/ dx is unique, if it exists; that is, there cannot be more than one number L that satisﬁes Deﬁnition 3.1.1. For brevity we will say “integrable” and “integral” when we mean “Riemann integrable” b a f .x/ dx exists is equivalent to saying that f is and “Riemann integral.” Saying that integrable on Œa; b. R Example 3.1.1 If then f .x/ n R D 1; a b; x n f .cj /.xj xj 1/ D .xj 1 j X D xj 1/: 1 j X D Most of the terms in the sum on the right cancel in pairs; that is, n 1 j X D .xj xj 1/ D .x1 x0/ C .x2 x1/ C C .xn xn 1/ .x1 x1/ .x2 x2/ C C C .xn 1 xn 1/ xn C x0 D xn b D D C x0 a: Thus, every Riemann sum of f over any partition of Œa; b equals b Example 3.1.2 Riemann sums for the function b a Z dx a: b D a, so are of the form Since xj cj 1 xj and .xj where f .x/ D x; a x b; D n cj .xj xj 1/: 1 j X D 1/=2 is the midpoint of Œxj xj 1 xj C 2 dj ; C xj C cj D j Substituting (3.1.4) into (3.1.3) yields j dj xj xj xj xj C 2 1 .xj xj 1x2 1 j X D j x2 1 dj .xj xj 1/ dj .xj xj 1/: (3.1.3) 1; xj , we can write (3.1.4) (3.1.5) (3.1.6) 116 Chapter 3 IntegralCalculusofFunctionsofOneVariable Because of cancellations like those in Example 3.1.1, n 1 j X D .x2 j x2 1/ j D b2 a2; so (3.1.6) can be rewritten as b2 a2 2 D n C 1 j X D dj .xj xj 1/: Hence, b2 a2 Xj 1 D P k 2 k a/: .b .xj dj j j xj 1/ k P 2 k n 1 Xj D .xj xj 1/ (see (3.1.5)) Therefore, every Riemann sum of f over a partition P of Œa; b satisﬁes b2 a2 2 ˇ ˇ ˇ ˇ Hence, < if < dx b2 a2 : 2 D The Integral as the Area Under a Curve An important application of the integral, indeed, the one invariably used to motivate its deﬁnition, is the computation of the area bounded by a curve y f .x/, the x-axis, and the lines x b (“the area under the curve”), as in Figure 3.1.1. a and x D D D y y = f (x) a b x Figure 3.1.1 Section 3.1 DeﬁnitionoftheIntegral 117 For simplicity, suppose that f .x/ > 0. Then f .cj /.xj 1 and height f .cj /, so the Riemann sum with base xj xj xj 1/ is the area of a rectangle n 1 j X D f .cj /.xj xj 1/ can be interpreted as the sum of the areas of rectangles related to the curve y shown in Figure 3.1.2. D f .x/, as y y = f (x) a c1 x1 c2 x2 c3 x3 c4 b x Figure 3.1.2 An apparently plausible argument, that the Riemann sums approximate the area under the curve more and more closely as the number of rectangles increases and the largest of b a f .x/ dx equals the their widths is made smaller, seems to support the assertion that area under the curve. This argument is useful as a motivation for Deﬁnition 3.1.1, which without it would seem mysterious. Nevertheless, the logic is incorrect, since it is based on the assumption that the area under the curve has been previously deﬁned in some other way. Although this is true for certain curves such as, for example, those consisting of line segments or circular arcs, it is not true in general. In fact, the area under a more complicated curve is deﬁned to be equal to the integral, if the integral exists. That this new deﬁnition is consistent with the old one, where the latter applies, is evidence that the integral provides a useful generalization of the deﬁnition of area. R Example 3.1.3 Let f .x/ 2 (Figure 3.1.3, page 118). The region under the curve consists of a square of unit area, surmounted by a triangle of area 1=2; thus, the area of the region is 3=2. From Example 3.1.2, x, 1 D x 2 x dx 1 2 .22 D 12/ 3 2 ; D 1 Z so the integral equals the area under the curve. 118 Chapter 3 IntegralCalculusofFunctionsofOneVariable y y = x 1 2 x Figure 3.1.3 y y = x 2 Example 3.1.4 If (Figure 3.1.4), then 1 2 x Figure 3.1.4 f .x/ x2; D 1 x 2 2 1 f .x/ dx 1 3 .23 D 13/ 7 3 D Z (Exercise 3.1.4), so we say that the area under the curve is 7=3. However, this is the deﬁnition of the area rather than a conﬁrmation of a previously known fact, as in Example 3.1.3. Section 3.1 DeﬁnitionoftheIntegral 119 Theorem 3.1.2 If f is unbounded on Œa; b; then f is not integrable on Œa; b: Proof We will show that if f is unbounded on Œa; b, P is any partition of Œa; b, and M > 0, then there are Riemann sums and 0 of f over P such that We leave it to you (Exercise 3.1.2) to complete the proof by showing from this that f cannot satisfy Deﬁnition 3.1.1. j 0 j M: (3.1.7) Let D n 1 j X D f .cj /.xj xj 1/ be a Riemann sum of f over a partition P of Œa; b. There must be an integer i in 1; 2; : : : ; n g such that f (3.1.8) f .c/ f .ci / j j xi M xi 1 for some c in Œxi 1xi , because if there were not so, we would have f .x/ j f .cj / j < M xj xj 1 ; xj 1 x xj ; 1 j n: Then f .x/ j D j f .cj / C j f .cj / j C j which implies that f .x/ f .cj / f .cj / f .x/ M xj xj 1 j j xj ; j C j x xj ; 1 1 f .cj / j n: j f .x/ j j max j 1 n j f .cj / j C M xj xj 1 ; a x b; contradicting the assumption that f is unbounded on Œa; b. Now suppose that c satisﬁes (3.1.8), and consider the Riemann sum 0 D n 1 Xj D f .c0j /.xj xj 1/ over the same partition P , where c0j D cj ; c; j j i; i: ¤ D 120 Chapter 3 IntegralCalculusofFunctionsofOneVariable Since (3.1.8) implies (3.1.7). j 0j D j f .c/ f .ci / j .xi xi 1/; Upper and Lower Integrals Because of Theorem 3.1.2, we consider only bounded functions throughout the rest of this section. To prove directly from Deﬁnition 3.1.1 that b a f .x/ dx exists, it is necessary to discover its value L in one way or another and to show that L has the properties required by the R deﬁnition. For a speciﬁc function it may happen that this can be done by straightforward calculation, as in Examples 3.1.1 and 3.1.2. However, this is not so if the objective is to ﬁnd b a f .x/ dx exists. The following approach avoids the general conditions which imply that difﬁculty of having to discover L in advance, without knowing whether it exists in the ﬁrst place, and requires only that we compare two numbers that must exist if f is bounded on Œa; b. We will see that b a f .x/ dx exists if and only if these two numbers are equal. R Deﬁnition 3.1.3 If f is bounded on Œa; b and P Œa; b, let R D f x0; x1; : : : ; xn g is a partition of Mj D sup x xj 1 f .x/ xj and The upper sum of f over P is mj xj 1 inf x xj D f .x/: S.P / n D 1 j X D Mj .xj xj 1/; and the upper integral of f over, Œa; b, denoted by f .x/ dx; b a Z is the inﬁmum of all upper sums. The lower sum of f over P is s.P / n D 1 j X D mj .xj xj 1/; and the lower integral of f over Œa; b, denoted by is the supremum of all lower sums. b a Z f .x/ dx; Section 3.1 DeﬁnitionoftheIntegral 121 If m f .x/ M for all x in Œa; b, then m.b a/ s.P / S.P / M.b a/ for every partition P ; thus, the set of upper sums of f over all partitions P of Œa; b is bounded, as is the set of lower sums. Therefore, Theorems 1.1.3 and 1.1.8 imply that b b a f .x/ dx exist, are unique, and satisfy the inequalities a f .x/ dx and R and R b m.b a/ m.b a/ a Z b a Z f .x/ dx M.b a/ f .x/ dx M.b a/: Theorem 3.1.4 Let f be bounded on Œa; b, and let P be a partition of Œa; b: Then (a) The upper sum S.P / of f over P is the supremum of the set of all Riemann sums of f over P: (b) The lower sum s.P / of f over P is the inﬁmum of the set of all Riemann sums of f over P: Proof (a) If P x0; x1; : : : ; xn D f g , then n S.P / D 1 j X D Mj .xj xj 1/; where Mj D sup x xj 1 f .x/: xj An arbitrary Riemann sum of f over P is of the form D n 1 j X D f .cj /.xj xj 1/; where xj xj . Since f .cj / Now let > 0 and choose cj in Œxj cj 1 f .cj / > Mj Mj , it follows that 1; xj so that n.xj xj 1/ ; 1 S.P /. n: j The Riemann sum produced in this way is D n 1 j X D f .cj /.xj xj 1/ > n Mj 1 j X D n.xj xj 1/ .xj xj 1/ D S.P / : / Now Theorem 1.1.3 implies that S.P / is the supremum of the set of Riemann sums of f over P . (b) Exercise 3.1.7. 122 Chapter 3 IntegralCalculusofFunctionsofOneVariable Example 3.1.5 Let f .x/ D 0 1 if x is irrational; if x is rational; and P nal and irrational numbers (Theorems 1.1.6 and 1.1.7), x0; x1; : : : ; xn D f g be a partition of Œa; b. Since every interval contains both ratio- Hence, and mj D 0 and Mj 1; 1 j n: D S..xj xj 1/ D b a .xj xj 1/ D 0: Since all upper sums equal b a and all lower sums equal 0, Deﬁnition 3.1.3 implies that b a Z f .x/ dx b D a and b a Z f .x/ dx 0: D Example 3.1.6 Let f be deﬁned on Œ1; 2 by f .x/ 0 if x is irrational and f .p=q/ 1=q if p and q are positive integers with no common factors (Exercise 2.2.7). If P x0; x1; : : : ; xn is any partition of Œ1; 2, then mj n, so s.P / 0; hence, 0 We now show that f .x/ dx 0: D 2 1 f .x/ dx 0 D (3.1.9) also. Since S.P / > 0 for every P , Deﬁnition 3.1.3 implies that so we need only show that 1 Z 2 2 f .x/ dx 0; f .x/ dx 0; which will follow if we show that no positive number is less than every upper sum. To this end, we observe that if 0 < < 2, then f .x/ =2 for only ﬁnitely many values of x in Œ1; 2. Let k be the number of such points and let P0 be a partition of Œ1; 2 such that P0 k k < 2k : (3.1.10) 2 1 Z Z 1 Z Consider the upper sum n S.P0/ D 1 j X D Section 3.1 DeﬁnitionoftheIntegral 123 Mj .xj xj 1/: 1 even for There are at most k values of j in this sum for which Mj these. The contribution of these terms to the sum is less
than k.=2k/ =2, because of (3.1.10). Since Mj < =2 for all other values of j , the sum of the other terms is less than =2, and Mj D 2 n .xj 1 j X D xj 1/ D 2 .xn x0/ D 2 .2 1/ D 2 : Therefore, S.P0/ < and, since can be chosen as small as we wish, no positive number is less than all upper sums. This proves (3.1.9). The motivation for Deﬁnition 3.1.3 can be seen by again considering the idea of area b, under a curve. Figure 3.1.5 shows the graph of a positive function y with Œa; b partitioned into four subintervals. f .x/, a D x y y = f (x) a x1 x2 x3 b x Figure 3.1.5 The upper and lower sums of f over this partition can be interpreted as the sums of the areas of the rectangles surmounted by the solid and dashed lines, respectively. This indicates that a sensible deﬁnition of area A under the curve must admit the inequalities for every partition P of Œa; b. Thus, A must be an upper bound for all lower sums and a lower bound for all upper sums of f over partitions of Œa; b. If s.P / A S.P / b a Z f .x/ dx D a Z b f .x/ dx; (3.1.11) 124 Chapter 3 IntegralCalculusofFunctionsofOneVariable there is only one number, the common value of the upper and lower integrals, with this property, and we deﬁne A to be that number; if (3.1.11) does not hold, then A is not deﬁned. We will see below that this deﬁnition of area is consistent with the deﬁnition stated earlier in terms of Riemann sums. Example 3.1.7 Returning to Example 3.1.3, consider the function If P D f x0; x1; : : : ; xn g f .x/ x; 1 x 2: is a partition of Œ1; 2, then, since f is increasing, D Hence, and Mj D f .xj / D xj and mj f .xj 1/ D xj 1: D xj .xj xj 1/ S..P / D 1 j X D xj 1.xj xj 1/: (3.1.12) (3.1.13) By writing we see from (3.1.12) that xj D n xj xj C 2 1 xj xj 2 1 ; C S.P / 1 2 1 2 D D .x2 j x2 1/ j 1 j X D .22 12/ 1 2 C xj 1/2: .xj 1 j X D 1 2 n .xj 1 j X D xj 1/2 (3.1.14) C n Since n 0 < 1 j X D (3.1.14) implies that .xj xj 1/2 P k k n .xj 1 j X D xj 1/ P .2 k D k 1/; 3 2 < S.P / 3 2 C k P 2 k : Since P k k can be made as small as we please, Deﬁnition 3.1.3 implies that A similar argument starting from (3.1.13) shows that b a f .x/ dx .P / < 3 2 ; Section 3.1 DeﬁnitionoftheIntegral 125 so b a Z f .x/ dx 3 2 : D Since the upper and lower integrals both equal 3=2, the area under the curve is 3=2 according to our new deﬁnition. This is consistent with the result in Example 3.1.3. The Riemann–Stieltjes Integral The Riemann–Stieltjes integral is an important generalization of the Riemann integral. We deﬁne it here, but conﬁne our study of it to the exercises in this and other sections of this chapter. Deﬁnition 3.1.5 Let f and g be deﬁned on Œa; b. We say that f is Riemann–Stieltjes integrable with respect to g on Œa; b if there is a number L with the following property: For every > 0, there is a ı > 0 such that n < ; (3.1.15) f .cj / g.xj / g.xj 1 provided only that P is a partition of Œa; b such that x0; x1; : : : ; xn g < ı and P k k xj 1 cj xj ; j D 1; 2; : : : ; n: In this case, we say that L is the Riemann–Stieltjes integral of f with respect to g over Œa; b, and write The sum f .x/ dg.x/ L: D b a Z n in (3.1.15) is a Riemann–Stieltjes sum of f with respect to g over the partition P . f .cj / g.xj / g.xj 1/ 1 j X D 3.1 Exercises Show that there cannot be more than one number L that satisﬁes Deﬁnition 3.1.1. (a) Prove: If 2 1 j of Œa; b with norms less than ı. b a f .x/ dx exists, then for every > 0, there is a ı > 0 such that < if 1 and 2 are Riemann sums of f over partitions P1 and P2 R j 1. 2. 126 Chapter 3 IntegralCalculusofFunctionsofOneVariable (b) Suppose that there is an M > 0 such that, for every ı > 0, there are Riemann sums 1 and 2 over a partition P of Œa; b with M . Use (a) to prove that f is not integrable over Œa; b. P k k < ı such that 1 j 2 j 3. 4. Suppose that and ı > 0, there is a partition P of Œa; b with over P that satisﬁes the inequality b a f .x/ dx exists and there is a number A such that, for every > 0 < ı and a Riemann sum of f P k < . Show that k R Prove directly from Deﬁnition 3.1.1 that j A j b a f .x/ dx A. D R b x2 dx b3 a3 : 3 D a Z Do not assume in advance that the integral exists. The proof of this is part of the be an arbitrary partition of Œa; b: Use problem. HINT: Let P the mean value theorem to show that x0; x2; : : : ; xn D f g b3 a3 xj xj 1/ for some points d1; . . . , dn; where xj arbitrary Riemann sums for f .x/ x2 over P: 1 < dj < xj . Then relate this sum to D 5. Generalize the proof of Exercise 3.1.4 to show directly from Deﬁnition 3.1.1 that b xm dx D a Z if m is an integer 0. bm 1 C m C 1 C am 1 6. Prove directly from Deﬁnition 3.1.1 that f .x/ is integrable on Œa; b if and only if f . x/ is integrable on Œ a, and, in this case, b; b f .x/ dx a Z a D b Z f . x/ dx: 7. Let f be bounded on Œa; b and let P be a partition of Œa; b. Prove: The lower sum s.P / of f over P is the inﬁmum of the set of all Riemann sums of f over P . 8. Let f be deﬁned on Œa; b and let P x0; x1; : : : ; xn be a partition of Œa; b. D f g (a) Prove: If f is continuous on Œa; b, then s.P / and S.P / are Riemann sums of f over P . (b) Name another class of functions for which the conclusion of (a) is valid. (c) Give an example where s.P / and S.P / are not Riemann sums of f over P . 9. Find 1 0 f .x/ dx and 1 0 f .x/ dx if Section 3.1 DeﬁnitionoftheIntegral 127 R (a) f .x/ 10. Given that x x if x is rational; R if x is irrational: D b a ex dx exists, evaluate it by using the formula R D 1 (b) f .xr 1/ ¤ 1 x if x is rational; if x is irrational: to calculate certain Riemann sums. HINT: See Exercise 3.1.3: b 0 sin x dx exists, evaluate it by using the identity 11. Given that R cos.j 1/ cos.j C 1/ D 2 sin sin j to calculate certain Riemann sums. HINT: See Exercise 3.1.3: b 0 cos x dx exists, evaluate it by using the identity 12. Given that R sin.j 1/ C sin.j 1/ D 2 sin cos j to calculate certain Riemann sums. HINT: See Exercise 3.1.3: 13. Show that if g.x/ b a f .x/ dx exists, in which case D C x c (c=constant), then R b a Z f .x/ dg.x/ f .x/ dx: D a Z b a f .x/ dg.x/ exists if and only if R b 14. Suppose that < a < d < c < and 1 1 g.x/ D g1; a < x < d; g2; d < x < b; (g1; g2 D constants), 15. 16. and let g.a/, g.b/, and g.d / be arbitrary. Suppose that f is deﬁned on Œa; b, continuous from the right at a and from the left at b, and continuous at d . Show that b a f .x/ dg.x/ exists, and ﬁnd its value. gm b < < a a0 < a1 < Suppose that R D 1 (constant) on .am p, and let g.a0/, g.a1/, . . . , g.ap / be arbitrary. 1; am/, 1 Suppose that f is deﬁned on Œa; b, continuous from the right at a and from the b a f .x/ dg.x/. HINT: See left at b, and continuous at a1, a2, . . . , ap Exercise 3.1.14: (a) Give an example where R b a f .x/ dg.x/ exists even though f is unbounded on Œa; b. (Thus, the analog of Theorem 3.1.2 does not hold for the Riemann– Stieltjes integral.) 1. Evaluate , let g.x/ < ap 1 D D m R (b) State and prove an analog of Theorem 3.1.2 for the case where g is increasing. 128 Chapter 3 IntegralCalculusofFunctionsofOneVariable 17. For the case where g is nondecreasing and f is bounded on Œa; b, deﬁne upper and lower Riemann–Stieltjes integrals in a way analogous to Deﬁnition 3.1.3. 3.2 EXISTENCE OF THE INTEGRAL The following lemma is the starting point for our study of the integrability of a bounded function f on a closed interval Œa; b. Lemma 3.2.1 Suppose that j and let P 0 be a partition of Œa; b obtained by adding r points to a partition P of Œa; b: Then j f .x/ M; a x b; and S.P / S.P 0/ s.P / s.P 0/ S.P / 2M r P k k s.P / 2M r P : k k C (3.2.1) x0; x1; : : : ; xn g D f (3.2.2) (3.2.3) Proof We will prove (3.2.2) and leave the proof of (3.2.3) to you (Exercise 3.2.1). First suppose that r x0; x1; : : : ; xn f uct Mj .xj S.P / D i , the prodf 1/ appears in both S.P / and S.P 0/ and cancels out of the difference S.P 0/. Therefore, if 1, so P 0 is obtained by adding one point c to the partition P 1 < c < xi for some i in 1; 2; : : : ; n g ; then xi . If j g xj D ¤ Mi1 D sup x xi1 f .x/ and Mi 2 D c c sup x xi f .x/; then S.P / S.P 0/ D D Since (3.2.1) implies that Mi .xi .Mi xi 1/ Mi1/.c Mi1.c xi 1/ C xi 1/ .Mi Mi 2.xi Mi 2/.xi c/ c/: (3.2.4) Mi 0 Mi r 2M; r 1; 2; D S.P / S.P 0/ 2M.xi xi 1/ 2M P : k k (3.2.4) implies that 0 This proves (3.2.2) for r 1. D Now suppose that r > 1 and P 0 is obtained by adding points c1, c2, . . . , cr to P . Let 1, let P .j / be the partition of Œa; b obtained by adding cj to P and, for j P .0/ P .j D 1/. Then the result just proved implies that 0 S.P .j 1// S.P .j // 2M k P .j 1/ ; k 1 j r: Adding these inequalities and taking account of cancellations yields Section 3.2 ExistenceoftheIntegral 129 0 Since P .0/ that S.P .0// S.P .r // P , P .r / D D P 0, and 2M. k P .k/ k P .0/ k C k P .k P .1/ 1/ k k k P .r 1/ /: (3.2.5) k 1, (3.2.5) implies k C C k for 1 k r which is equivalent to (3.2.2). 0 S.P / S.P 0/ 2M r P ; k k Theorem 3.2.2 If f is bounded on Œa; b; then b a Z f .x/ dx a Z b f .x/ dx: (3.2.6) Proof Suppose that P1 and P2 are partitions of Œa; b and P 0 is a reﬁnement of both. Letting P P2 in (3.2.2) shows that P1 in (3.2.3) and P D D s.P 0/ s.P1/ and S.P 0/ S.P2/: Since s.P 0/ bound for the set of all upper sums. Since that S.P 0/, this implies that s.P1/ S.P2/. Thus, every lower sum is a lower b a f .x/ dx is the inﬁmum of this set, it follows R b s.P1/ a Z f .x/ dx for every partition P1 of Œa; b. This means that of all lower sums. Since b a f .x/ dx is the supremum of this set, this implies (3.2.6). R b a f .x/ dx is an upper bound for the set R Theorem 3.2.3 If f is integrable on Œa; b; then b a Z f .x/ dx D a Z b f .x/ dx D a Z b f .x/ dx: Proof We prove that b a f .x/ dx b a f .x/ dx and leave it to you to show that b a f .x/ dx D b a f .x/ dx (Exercise 3.2.2). Suppose that P is a partition of Œa; b and is a Riemann sum of f over P . Since x/ dx b f .x/ dx b D Z a a Z f .x/ dx S.P / ! C .S.P / / b C a Z f .x/ dx ; ! 130 Chapter 3 IntegralCalculusofFunctionsofOneVariable the triangle
inequality implies that b a Z f .x/ dx a Z b f .x/ dx b a f .x/ dx Cˇ ˇ ˇ ˇ Now suppose that > 0. From Deﬁnition 3.1.3, there is a partition P0 of Œa; b such that ˇ Z a S.P / ˇ ˇ ˇ ˇ : f .x/ dx ˇ ˇ ˇ ˇ ˇ ˇ C j S.P / j (3.2.7) b f .x/ dx b S.P0/ < f .x/ dx a 3 : C Z From Deﬁnition 3.1.1, there is a ı > 0 such that a Z b f .x/ dx . Now suppose that if by Lemma 3.2.1, (3.2.8) implies that k k k < ı and P is a reﬁnement of P0. Since S.P / b a Z f .x/ dx so b S.P / < f .x/ dx 3 ; C a Z b S.P / f .x/ dx < 3 ˇ ˇ ˇ in addition to (3.2.9). Now (3.2.7), (3.2.9), and (3.2.10) imply that 3.2.8) (3.2.9) S.P0/ (3.2.10) (3.2.11) b a Z f .x/ dx a Z b f .x/ dx < 2 3 C j S. for every Riemann sum of f over P . Since S.P / is the supremum of these Riemann sums (Theorem 3.1.4), we may choose so that Now (3.2.11) implies that S.P / j < j 3 : Since is an arbitrary positive number, it follows that f .x/ dx b a Z b a Z f .x/ dx < : x/ dx D a Z b f .x/ dx: Section 3.2 ExistenceoftheIntegral 131 Lemma 3.2.4 If f is bounded on Œa; b and > 0; there is a ı > 0 such that b f .x/ dx a Z b a Z f .x/ dx b S.P / < f .x/ dx a Z b s.P / > f .x/ dx a Z C (3.2.12) and if P k k < ı. Proof We show that (3.2.12) holds if proof to you (Exercise 3.2.3). P k k is sufﬁciently small, and leave the rest of the The ﬁrst inequality in (3.2.12) follows immediately from Deﬁnition 3.1.3. To establish b. From Deﬁnition 3.1.3, there f .x/ the second inequality, suppose that x0; x1; : : : ; xr is a partition P0 x K if a of Œa; b such that j D f j C 1 g b S.P0/ < f .x/ dx a Z 2 : C (3.2.13) If P is any partition of Œa; b, let P 0 be constructed from the partition points of P0 and P . Then S.P 0/ S.P0/; (3.2.14) by Lemma 3.2.1. Since P 0 is obtained by adding at most r points to P , Lemma 3.2.1 implies that Now (3.2.13), (3.2.14), and (3.2.15) imply that S.P 0/ S.P / 2Kr P : k k (3.2.15) S.P / < S.P 0/ S.P0/ b 2Kr 2Kr P P k k C C f .x/ dx C a Z k k 2 C 2Kr P : k k Therefore, (3.2.12) holds if < ı P k k D 4Kr : 132 Chapter 3 IntegralCalculusofFunctionsofOneVariable Theorem 3.2.5 If f is bounded on Œa; b and f .x/ dx b a Z b D a Z f .x/ dx L; D then f is integrable on Œa; b and b a Z f .x/ dx L: D (3.2.16) (3.2.17) Section 3.2 ExistenceoftheIntegral 133 Proof If > 0, there is a ı > 0 such that b a Z f .x/ dx < s.P / b S.P / < f .x/ dx a Z C (3.2.18) if P k k < ı (Lemma 3.2.4). If is a Riemann sum of f over P , then so (3.2.16) and (3.2.18) imply that s.P / S.P /; < < L L C if P < ı. Now Deﬁnition 3.1.1 implies (3.2.17). k Theorems 3.2.3 and 3.2.5 imply the following theorem. k Theorem 3.2.6 A bounded function f is integrable on Œa; b if and only if b a Z f .x/ dx D a Z b f .x/ dx: The next theorem translates this into a test that can be conveniently applied. Theorem 3.2.7 If f is bounded on Œa; b; then f is integrable on Œa; b if and only if for each > 0 there is a partition P of Œa; b for which S.P / s.P / < : (3.2.19) Proof We leave it to you (Exercise 3.2.4) to show that if holds for tegrability. To show that it is sufﬁcient, we observe that since b a f .x/ dx exists, then (3.2.19) sufﬁciently small. This implies that the stated condition is necessary for in- P k k R s.P / a Z for all P , (3.2.19) implies that b b f .x/ dx b a Z f .x/ dx S.P / f .x/ dx 0 a Z b a Z f .x/ dx < : Since can be any positive number, this implies that b a Z f .x/ dx D a Z b f .x/ dx: Therefore, b a f .x/ dx exists, by Theorem 3.2.5. The next two theorems are important applications of Theorem 3.2.7. R 134 Chapter 3 IntegralCalculusofFunctionsofOneVariable Theorem 3.2.8 If f is continuous on Œa; b; then f is integrable on Œa; b. Proof Let P there are points cj and c0j in Œxj x0; x1; : : : ; xn D f g 1; xj such that be a partition of Œa; b. Since f is continuous on Œa; b, f .cj / Mj D D sup x xj 1 f .x/ xj and (Theorem 2.2.9). Therefore, f .c0j / mj D D xj 1 inf x xj f .x/ S.P / s.P / D n f .cj / 1 j X D f .c0j / .xj xj 1/: (3.2.20) Since f is uniformly continuous on Œa; b (Theorem 2.2.12), there is for each > 0 a ı > 0 such that f .x0/ < ı. If j x0j f .x, then if x and x0 are in Œa; b and x j cj j c0j j < ı and, from (3.2.20), S.P / s.P / < b n a .xj 1 j X D xj 1/ D : Hence, f is integrable on Œa; b, by Theorem 3.2.7. Theorem 3.2.9 If f is monotonic on Œa; b; then f is integrable on Œa; b. Proof Let P D f x0; x1; : : : ; xn g f .xj / be a partition of Œa; b. Since f is nondecreasing, Mj D D sup x xj 1 f .x/ xj and Hence, S.P / f .xj 1/ D mj D xj 1 inf x xj f .x/: n .f .xj / s.P / D 1 j X D and f .xj / f .xj 1//.xj xj 1/: Since 0 < xj xj 1 P k k f .xj 1/ 0, n S.P / s.f .xj / 1 j X D .f .b/ f .a//: f .xj 1// Therefore, k so f is integrable on Œa; b, by Theorem 3.2.7. S.P / s.P / < if Section 3.2 ExistenceoftheIntegral 135 .f .b/ P k f .a// < ; The proof for nonincreasing f is similar. We will also use Theorem 3.2.7 in the next section to establish properties of the integral. In Section 3.5 we will study more general conditions for integrability. 3.2 Exercises 1. Complete the proof of Lemma 3.2.1 by verifying Eqn. (3.2.3). 2. Show that if f is integrable on Œa; b, then b a Z f .x/ dx D a Z b f .x/ dx: 3. Prove: If f is bounded on Œa; b, there is for each > 0 a ı > 0 such that 4. 5. 6. f .x/ dx b a Z b a Z f .x/ dx < s.P / < ı. if P k k Prove: If f is integrable on Œa; b and > 0, then S.P / sufﬁciently small. HINT: Use Theorem 3.1.4: Suppose that f is integrable and g is bounded on Œa; b, and g differs from f only at points in a set H with the following property: For each > 0, H can be covered by a ﬁnite number of closed subintervals of Œa; b, the sum of whose lengths is less than . Show that g is integrable on Œa; b and that s.P / < if P is k k b a g.x/ dx D b a f .x/ dx: Z Z HINT: Use Exercise 3.1.3: Suppose that g is bounded on Œ˛; ˇ, and let Q a ﬁxed partition of Œ˛; ˇ. Prove: ˛ W D v0 < v1 < (a) ˇ ˛ Z g.u/ du L v` D v`1 1 Z X` D g.u/ du I (b) g.u/ du ˇ ˛ Z < vL ˇ be D v` L g.u/ du: D v`1 1 Z X` D 7. A function f is of bounded variation on Œa; b if there is a number K such that n f .aj / f .aj 1/ K ˇ ˇ whenever a < an is the total variation of f on Œa; b.) a0 < a1 < D 1 j X D b. (The smallest number with this property ˇ ˇ D 136 Chapter 3 IntegralCalculusofFunctionsofOneVariable (a) Prove: If f is of bounded variation on Œa; b, then f is bounded on Œa; b. (b) Prove: If f is of bounded variation on Œa; b, then f is integrable on Œa; b. HINT: Use Theorems 3.1.4 and 3.2.7: 8. Let P and xj n x0; x1; : : : ; xn D f 1 cj xj , j g D be a partition of Œa; b, c0 1, 2, . . . , n. Verify that x0 D D a, cn 1 C xn b, D D g.cj /Œf .xj / f .xj 1/ g.b/f .b/ g.a/f .a/ D 1 Xj D Use this to prove that if 0 Xj D b a f .x/ dg.x/ exists, then so does n f .xj /Œg.cj 1/ g.cj /: C b a g.x/ df .x/, and R g.x/ df .x/ b a Z f .b/g.b/ f .a/g.a/ D a Z b R f .x/ dg.x/: (This is the integration by parts formula for Riemann–Stieltjes integrals.) 9. Let f be continuous and g be of bounded variation (Exercise 3.2.7) on Œa; b. (a) Show that if > 0, there is a ı > 0 such that < =2 if and 0 0j are Riemann–Stieltjes sums of f with respect to g over partitions P and P 0 < ı. HINT: Use Theoof Œa; b, where P 0 is a reﬁnement of P and rem 2.2.12: P k k j (b) Let ı be as chosen in (a). Suppose that 1 and 2 are Riemann–Stieltjes sums of f with respect to g over any partitions P1 and P2 of Œa; b with norm less than ı. Show that < . 1 j 2 j (c) If ı > 0, let L.ı/ be the supremum of all Riemann–Stieltjes sums of f with respect to g over partitions of Œa; b with norms less than ı. Show that L.ı/ is ﬁnite. Then show that L L.ı/ exists. HINT: Use Theorem 2.1.9: limı (d) Show that D b a f .x/ dg.x/ L. D 0 ! C 10. Show that Œa; b. HINT: See Exercises 3.2.8 and 3.2.9: b a f .x/ dg.x/ exists if f is of bounded variation and g is continuous on R R 3.3 PROPERTIES OF THE INTEGRAL We now use the results of Sections 3.1 and 3.2 to establish the properties of the integral. You are probably familiar with most of these properties, but not with their proofs. Theorem 3.3.1 If f and g are integrable on Œa; b; then so is f g; and C b .f a Z g/.x/ dx C D a Z b f .x/ dx C a Z b g.x/ dx: Section 3.3 PropertiesoftheIntegral 137 Proof Any Riemann sum of f be written as C g over a partition P x0; x1; : : : ; xn g D f of Œa; b can Œf .cj / C g.cj /.xj xj 1/ xj 1/ C n 1 j X D g.cj /.xj xj 1/ f .cj /.xj where f and g are Riemann sums for f and g. Deﬁnition 3.1.1 implies that if > 0 there are positive numbers ı1 and ı2 such that b f .x/ dx a Z g b a g.x/ dx min.ı1; ı2/, then and If x/ dx .x/ dx ˇ ˇ ˇ ˇ ˇ if P k k < ı1 if P k k < ı2x/ dx g ! C a Z b b g.x/ dx f f .x/ dx .x/ dx ˇ ˇ ˇ ˇ ˇ 2 C so the conclusion follows from Deﬁnition 3.1.1. < The next theorem also follows from Deﬁnition 3.1.1 (Exercise 3.3.1). Theorem 3.3.2 If f is integrable on Œa; b and c is a constant; then cf is integrable on Œa; b and b a Z cf .x/ dx b c D a Z f .x/ dx: Theorems 3.3.1 and 3.3.2 and induction yield the following result (Exercise 3.3.2). Theorem 3.3.3 If f1; f2; . . . ; fn are integrable on Œa; b and c1; c2; . . . ; cn are constants; then c1f1 cnfn is integrable on Œa; b and c2f2 C C C b a Z .c1f1 c2f2 C C C cnfn/.x/ dx b c1 D a Z f1.x/ dx b b c2 C a Z C C cn fn.x/ dx: a Z f2.x/ dx 138 Chapter 3 IntegralCalculusofFunctionsofOneVariable Theorem 3.3.4 If f and g are integrable on Œa; b and f .x/ then b b f .x/ dx a Z a Z g.x/ dx: g.x/ for a x b; (3.3.1) Proof Since g.x/ nonnegative. Therefore, f .x/ 0, every lower sum of g b .g.x/ f .x// dx f over any partition of Œa; b is 0: Hence.x/ dx f .x/ dx b D a Z b D a Z .g.x/ .g.x/ f .x// dx f .x// dx 0; (3.3.2) which yields (3.3.1). (The ﬁrst equality in (3.3.2) follows from Theorems 3.3.1 and 3.3.2; the second, from Theorem 3.2.3.) Theorem 3.3.5 If f is integrable on Œa; b; then so is , and f j j b a Z f .x/ dx b a j Z f .x/ j dx: (3.3.3) Proof Let P be a partition of Œa; b and deﬁne Mj mj M j mj D D D D sup inf sup ˚ inf f .x/ f .x/ ˚ xj xj ˇ ˇ f .x/ ˇ j j ˇ f .x/ ˚ j j ˚ f .x/ f .x/ mj : xj xj ˇ ˇ ˇ ˇ j j ; xj xj x x 1 1 1 1 x x ; xj xj ; :

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